Printing a two-dimensional array in a spiral order

Question

How to print a two-dimensional 5 × 5 array in a spiral order?

Is there any formula so that I can print an array of any size in spiral order?

Answer 1

The idea is to consider the matrix as a sequence of layers, upper right layers and lower left layers. To print the matrix in a spiral, we can clear the layers of this matrix, print the cleaned part and recursively call the print on the left above the part. Recursion ends when we no longer have print layers.

Input Matrix:

1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 1 

After peeling the upper right layer:

  1 2 3 4 8 5 6 7 2 9 0 1 6 3 4 5 1 7 8 9 

After detaching the lower left layer from the submatrix:

  6 7 5 0 1 9 4 5 3 7 8 9 

After cleaning the upper right layer of the sub-matrix:

  6 7 1 0 5 4 

After cleaning the bottom left layer of the sub-matrix:

  0 4 

The recursion ends.

---

With function:

 // function to print the top-right peel of the matrix and // recursively call the print bottom-left on the submatrix. void printTopRight(int a[][COL], int x1, int y1, int x2, int y2) { int i = 0, j = 0; // print values in the row. for(i = x1; i<=x2; i++) { printf("%d ", a[y1][i]); } // print values in the column. for(j = y1 + 1; j <= y2; j++) { printf("%d ", a[j][x2]); } // see if more layers need to be printed. if(x2-x1 > 0) { // if yes recursively call the function to // print the bottom left of the sub matrix. printBottomLeft(a, x1, y1 + 1, x2-1, y2); } } // function to print the bottom-left peel of the matrix and // recursively call the print top-right on the submatrix. void printBottomLeft(int a[][COL], int x1, int y1, int x2, int y2) { int i = 0, j = 0; // print the values in the row in reverse order. for(i = x2; i>=x1; i--) { printf("%d ", a[y2][i]); } // print the values in the col in reverse order. for(j = y2 - 1; j >= y1; j--) { printf("%d ", a[j][x1]); } // see if more layers need to be printed. if(x2-x1 > 0) { // if yes recursively call the function to // print the top right of the sub matrix. printTopRight(a, x1+1, y1, x2, y2-1); } } void printSpiral(int arr[][COL]) { printTopRight(arr,0,0,COL-1,ROW-1); printf("\n"); } 

Answer 2

• Top row in half
• Rearrange and flip upside down (same as rotating 90 degrees counterclockwise)
• Go to 1

Python Code:

 import itertools arr = [[1,2,3,4], [12,13,14,5], [11,16,15,6], [10,9,8,7]] def transpose_and_yield_top(arr): while arr: yield arr[0] arr = list(reversed(zip(*arr[1:]))) print list(itertools.chain(*transpose_and_yield_top(arr))) 

Answer 3

I see that no one uses only one for loop and without recursion in the code, so I want to contribute.

The idea is this:

Imagine that at point (0,0) there is a tortoise, that is, the upper left corner facing east (right)

He will continue to go forward and every time he sees a sign, the turtle will turn right

So, if we place the turtle at the point (0,0) facing right, and if we place the signs in the appropriate places, the turtle will move in a spiral.

Now the problem is: "Where to put the signs?"

Let's see where we should put the signs (marked with a # and numbers on O):

 For a grid that looks like this:
 OOOO
 OOOO
 OOOO
 OOOO

 We put the signs like this:
 OOO #
 # O # O
 O # # O
 # OO #

 For a grid that looks like this:
 OOO
 OOO
 OOO
 OOO

 We put the signs like this:
 OO #
 # # O
 O # o
 # O #

 And for a grid that looks like this:
 OOOOOOO
 OOOOOOO
 OOOOOOO
 OOOOOOO
 OOOOOOO

 We put the signs like this:
 OOOOOO #
 # OOOO # O
 O # OO # OO
 O # OOO # O
 # OOOOO #

We can see that if the point is not in the upper left side, the signs are at points where the distances to the nearest horizontal border and the nearest vertical border are the same , while for the upper left side the distance to the upper border is more than the distance to left border , with priority given to the upper right if the point is horizontally centered, and the top to the left if the point is centered vertically.

This can be easily implemented with a simple function, taking a minimum ( curRow and height-1-curRow ), then a minimum ( curCol and width-1-curCol ) and compare if they are the same. But we need to consider the upper left case, i.e. When the minimum is equal to curRow and curCol . In this case, we accordingly reduce the vertical distance.

Here is the C code:

 #include <stdio.h> int shouldTurn(int row, int col, int height, int width){ int same = 1; if(row > height-1-row) row = height-1-row, same = 0; // Give precedence to top-left over bottom-left if(col >= width-1-col) col = width-1-col, same = 0; // Give precedence to top-right over top-left row -= same; // When the row and col doesn't change, this will reduce row by 1 if(row==col) return 1; return 0; } int directions[4][2] = {{0,1},{1,0},{0,-1},{-1,0}}; void printSpiral(int arr[4][4], int height, int width){ int directionIdx=0, i=0; int curRow=0, curCol=0; for(i=0; i<height*width; i++){ printf("%d ",arr[curRow][curCol]); if(shouldTurn(curRow, curCol, height, width)){ directionIdx = (directionIdx+1)%4; } curRow += directions[directionIdx][0]; curCol += directions[directionIdx][1]; } printf("\n"); } int main(){ int arr[4][4]= {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}}; printSpiral(arr, 4, 4); printSpiral(arr, 3, 4); } 

What outputs:

 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10
 1 2 3 4 8 12 11 10 9 5 6 7

Answer 4

Here are three interesting ways:

• Reading in a spiral way can be considered as a snake moving to the border and including hitting the border or itself (I find it elegant and most effective, being the only loop of N iterations)

   ar = [ [ 0, 1, 2, 3, 4], [15, 16, 17, 18, 5], [14, 23, 24, 19, 6], [13, 22, 21, 20, 7], [12, 11, 10, 9, 8]] def print_spiral(ar): """ assuming a rect array """ rows, cols = len(ar), len(ar[0]) r, c = 0, -1 # start here nextturn = stepsx = cols # move so many steps stepsy = rows-1 inc_c, inc_r = 1, 0 # at each step move this much turns = 0 # how many times our snake had turned for i in range(rows*cols): c += inc_c r += inc_r print ar[r][c], if i == nextturn-1: turns += 1 # at each turn reduce how many steps we go next if turns%2==0: nextturn += stepsx stepsy -= 1 else: nextturn += stepsy stepsx -= 1 # change directions inc_c, inc_r = -inc_r, inc_c print_spiral(ar) 

exit:

 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 

2. A recursive approach is to print the outer layer and call the same function for the inner rectangle, for example.

    def print_spiral(ar, sr=0, sc=0, er=None, ec=None): er = er or len(ar)-1 ec = ec or len(ar[0])-1 if sr > er or sc > ec: print return # print the outer layer top, bottom, left, right = [], [], [], [] for c in range(sc,ec+1): top.append(ar[sr][c]) if sr != er: bottom.append(ar[er][ec-(c-sc)]) for r in range(sr+1,er): right.append(ar[r][ec]) if ec != sc: left.append(ar[er-(r-sr)][sc]) print " ".join([str(a) for a in top + right + bottom + left]), # peel next layer of onion print_spiral(ar, sr+1, sc+1, er-1, ec-1) 

Finally, here is a small snippet to do this, but not effective, but fun :), basically it prints the top row and rotates the entire rectangle counterclockwise and repeats

 def print_spiral(ar): if not ar: return print " ".join(str(a) for a in ar[0]), ar = zip(*[ reversed(row) for row in ar[1:]]) print_spiral(ar) 

Answer 5

This program works for any n * n matrix.

 public class circ { public void get_circ_arr (int n,int [][] a) { int z=n; { for (int i=0;i<n;i++) { for (int l=z-1-i;l>=i;l--) { int k=i; System.out.printf("%d",a[k][l]); } for (int j=i+1;j<=z-1-i;j++) { int k=i; { System.out.printf("%d",a[j][k]); } } for (int j=i+1;j<=zi-1;j++) { int k=z-1-i; { System.out.printf("%d",a[k][j]); } } for (int j=z-2-i;j>=i+1;j--) { int k=zi-1; { System.out.printf("%d",a[j][k]); } } } } } } 

Hope this helps

Answer 6

I was obsessed with this problem when I was learning Ruby. This was the best I could do:

 def spiral(matrix) matrix.empty? ? [] : matrix.shift + spiral(matrix.transpose.reverse) end 

You can check out some of my other solutions by returning to the revisions of this chapter . In addition, if you follow the link from which I received a response, you will find other smart solutions. A really interesting problem that can be solved in several elegant ways - especially in Ruby.

Answer 7

JavaScript solution:

 var printSpiral = function (matrix) { var i; var top = 0; var left = 0; var bottom = matrix.length; var right = matrix[0].length; while (top < bottom && left < right) { //print top for (i = left; i < right; i += 1) { console.log(matrix[top][i]); } top++; //print right column for (i = top; i < bottom; i += 1) { console.log(matrix[i][right - 1]); } right--; if (top < bottom) { //print bottom for (i = right - 1; i >= left; i -= 1) { console.log(matrix[bottom - 1][i]); } bottom--; } if (left < right) { //print left column for (i = bottom - 1; i >= top; i -= 1) { console.log(matrix[i][left]); } left++; } } }; 

Answer 8

One solution includes directions on the right, left, up, down and their corresponding limits (indices). After printing the first line and changing the direction (right) down, the line is discarded by increasing the upper limit. After the last column is printed, and the direction changes to the left, the column will be discarded by reducing the limit of the right hand ... Details can be seen in a clear C-code.

 #include <stdio.h> #define N_ROWS 5 #define N_COLS 3 void print_spiral(int a[N_ROWS][N_COLS]) { enum {up, down, left, right} direction = right; int up_limit = 0, down_limit = N_ROWS - 1, left_limit = 0, right_limit = N_COLS - 1, downcount = N_ROWS * N_COLS, row = 0, col = 0; while(printf("%d ", a[row][col]) && --downcount) if(direction == right) { if(++col > right_limit) { --col; direction = down; ++up_limit; ++row; } } else if(direction == down) { if(++row > down_limit) { --row; direction = left; --right_limit; --col; } } else if(direction == left) { if(--col < left_limit) { ++col; direction = up; --down_limit; --row; } } else /* direction == up */ if(--row < up_limit) { ++row; direction = right; ++left_limit; ++col; } } void main() { int a[N_ROWS][N_COLS] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}; print_spiral(a); } 

Link for testing and downloading

.

Answer 9

The two-dimensional matrix N * N is a square matrix

Idea:

We need to go in four different directions to go in a spiral. We must cross the inner matrix after one layer of the spiral ends. So, in total, we need 5 loops, 4 loops for passing in a spiral and 1 loop for passing through layers.

 public void printSpiralForm(int[][] a, int length) { for( int i = 0 , j = length-1 ; i < j ; i++ , j-- ) { for( int k = i ; k < j ; k++ ) { System.out.print( a[i][k] + " " ) ; } for( int k = i ; k < j ; k++ ) { System.out.print(a[k][j] + " "); } for( int k = j ; k > i ; k-- ) { System.out.print(a[j][k] + " ") ; } for( int k = j ; k > i ; k-- ) { System.out.print( a[k][i] + " " ) ; } } if ( length % 2 == 1 ) { System.out.println( a[ length/2 ][ length/2 ] ) ; } } 

Answer 10

Given the character matrix, implement a method that prints all characters in the following order: first the outer circle, then the next, etc.

 public static void printMatrixInSpiral(int[][] mat){ if(mat.length == 0|| mat[0].length == 0){ /* empty matrix */ return; } StringBuffer str = new StringBuffer(); int counter = mat.length * mat[0].length; int startRow = 0; int endRow = mat.length-1; int startCol = 0; int endCol = mat[0].length-1; boolean moveCol = true; boolean leftToRight = true; boolean upDown = true; while(counter>0){ if(moveCol){ if(leftToRight){ /* printing entire row left to right */ for(int i = startCol; i <= endCol ; i++){ str.append(mat[startRow][i]); counter--; } leftToRight = false; moveCol = false; startRow++; } else{ /* printing entire row right to left */ for(int i = endCol ; i >= startCol ; i--){ str.append(mat[endRow][i]); counter--; } leftToRight = true; moveCol = false; endRow--; } } else { if(upDown){ /* printing column up down */ for(int i = startRow ; i <= endRow ; i++){ str.append(mat[i][endCol]); counter--; } upDown = false; moveCol = true; endCol--; } else { /* printing entire col down up */ for(int i = endRow ; i >= startRow ; i--){ str.append(mat[i][startCol]); counter--; } upDown = true; moveCol = true; startCol++; } } } System.out.println(str.toString()); } 

Answer 11

Just keep it simple →

 public class spiralMatrix { public static void printMatrix(int[][] matrix, int rows, int col) { int rowStart=0; int rowEnd=rows-1; int colStart=0; int colEnd=col-1; while(colStart<=colEnd && rowStart<=rowEnd) { for(int i=colStart;i<colEnd;i++) System.out.println(matrix[rowStart][i]); for(int i=rowStart;i<rowEnd;i++) System.out.println(matrix[i][colEnd]); for(int i=colEnd;i>colStart;i--) System.out.println(matrix[rowEnd][i]); for(int i=rowEnd;i>rowStart;i--) System.out.println(matrix[i][colStart]); rowStart++; colEnd--; rowEnd--; colStart++; } } public static void main(String[] args){ int[][] array={{1,2,3,4},{5,6,7,8}}; printMatrix(array,2,4); } 

}

Answer 12

This is my implementation:

 public static void printMatrix(int matrix[][], int M, int N){ int level = 0; int min = (M < N) ? M:N; System.out.println(); while(level <= min/2){ for(int j = level; j < N - level - 1; j++){ System.out.print(matrix[level][j] + "\t"); } for(int i = level; i < M - level - 1; i++) { System.out.print(matrix[i][N - level - 1] + "\t"); } for(int j = N - level - 1; j > level; j--){ System.out.print(matrix[M - level - 1][j] + "\t"); } for(int i = M - level - 1; i > level; i-- ){ System.out.print(matrix[i][level] + "\t"); } level++; } } 

Answer 13

int N = Integer.parseInt (args [0]);

  // create N-by-N array of integers 1 through N int[][] a = new int[N][N]; for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) a[i][j] = 1 + N*i + j; // spiral for (int i = N-1, j = 0; i > 0; i--, j++) { for (int k = j; k < i; k++) System.out.println(a[j][k]); for (int k = j; k < i; k++) System.out.println(a[k][i]); for (int k = i; k > j; k--) System.out.println(a[i][k]); for (int k = i; k > j; k--) System.out.println(a[k][j]); } // special case for middle element if N is odd if (N % 2 == 1) System.out.println(a[(N-1)/2][(N-1)/2]); } 

}

Answer 14

Slash Top Row → Transpose → Flip → Repeat.

 void slashTransposeFlip(int[][] m){ if( m.length * m[0].length == 1){ //only one element left System.out.print(m[0][0]); }else{ //print the top row for(int a:m[0]){System.out.print(a+" ");} //slash the top row from the matrix. int[][] n = Arrays.copyOfRange(m,1,m.length); int[][] temp = n; int rows = temp.length; int columns = temp[0].length; //invert rows and columns and create new array n = new int[columns][rows]; //transpose for(int x=0;x<rows;x++) for(int y=0;y<columns;y++) n[y][x] = temp[x][y]; //flipping time for (int i = 0; i < n.length / 2; i++) { int[] t = n[i]; n[i] = n[n.length - 1 - i]; n[n.length - 1 - i] = t; } //recursively call again the reduced matrix. slashTransposeFlip(n); } } 

Answer 15

Difficulty: Single traverse O(n)

Let me add my single loop response with O(n) complexity. I noticed that during the left and right left movement of the matrix there is an increase and decrease by one, respectively, in the row index. Similarly, for the top and bottom, the stroke increases and decreases by n_cols . So I made an algorithm for this. For example, given the (3x5) matrix with entries, the main row indices are displayed as follows: 1,2,3,4,5,10,15,14,13,12,11,6,7,8,9 .

  ------->(+1) ^ 1 2 3 4 5 | (+n_cols) | 6 7 8 9 10 | (-n_cols) | 11 12 13 14 15 (-1)<------- 

Code Solution:

 #include <iostream> using namespace std; int main() { // your code goes here bool leftToRight=true, topToBottom=false, rightToLeft=false, bottomToTop=false; int idx=0; int n_rows = 3; int n_cols = 5; int cnt_h = n_cols, cnt_v = n_rows, cnt=0; int iter=1; for (int i=0; i <= n_rows*n_cols + (n_rows - 1)*(n_cols - 1)/2; i++){ iter++; if(leftToRight){ if(cnt >= cnt_h){ cnt_h--; cnt=0; leftToRight = false; topToBottom = true; //cout << "Iter: "<< iter << " break_leftToRight"<<endl; }else{ cnt++; idx++; //cout << "Iter: "<< iter <<" idx: " << idx << " cnt: "<< cnt << " cnt_h: "<< cnt_h<< endl; cout<< idx << endl; } }else if(topToBottom){ if(cnt >= cnt_v-1){ cnt_v--; cnt=0; leftToRight = false; topToBottom = false; rightToLeft=true; //cout << "Iter: "<< iter << " break_topToBottom"<<endl; }else{ cnt++; idx+=n_cols; //cout << "Iter: "<< iter << " idx: " << idx << " cnt: "<< cnt << " cnt_v: "<< cnt_h<< endl; cout << idx <<endl; } }else if(rightToLeft){ if(cnt >= cnt_h){ cnt_h--; cnt=0; leftToRight = false; topToBottom = false; rightToLeft=false; bottomToTop=true; //cout << "Iter: "<< iter << " break_rightToLeft"<<endl; //cout<< idx << endl; }else{ cnt++; idx--; //cout << "Iter: "<< iter << " idx: " << idx << " cnt: "<< cnt << " cnt_h: "<< cnt_h<< endl; cout << idx <<endl; } }else if(bottomToTop){ if(cnt >= cnt_v-1){ cnt_v--; cnt=0; leftToRight = true; topToBottom = false; rightToLeft=false; bottomToTop=false; //cout << "Iter: "<< iter << " break_bottomToTop"<<endl; }else{ cnt++; idx-=n_cols; //cout << "Iter: "<< iter << " idx: " << idx << " cnt: "<< cnt << " cnt_v: "<< cnt_h<< endl; cout<< idx << endl; } } //cout << i << endl; } return 0; } 

Answer 16

To print a two-dimensional matrix, consider the matrix as a composition of rectangles and / or lines, where the smaller rectangle is set to the larger one, take the border of the matrix, which will form a rectangle that will be printed starting from the element up-left every time in each layer; as soon as you do this, go inside for the next layer of the smaller rectangle, in case I don't have a rectangle, then it should be a line for printing, horizontal or vertical. I pasted the code with an example matrix, HTH.

 #include <stdio.h> int a[2][4] = { 1, 2 ,3, 44, 8, 9 ,4, 55 }; void print(int, int, int, int); int main() { int row1, col1, row2, col2; row1=0; col1=0; row2=1; col2=3; while(row2>=row1 && col2>=col1) { print(row1, col1, row2, col2); row1++; col1++; row2--; col2--; } return 0; } void print(int row1, int col1, int row2, int col2) { int i=row1; int j=col1; /* This is when single horizontal line needs to be printed */ if( row1==row2 && col1!=col2) { for(j=col1; j<=col2; j++) printf("%d ", a[i][j]); return; } /* This is when single vertical line needs to be printed */ if( col1==col2 && row1!=row2) { for(i=row1; j<=row2; i++) printf("%d ", a[i][j]); return; } /* This is reached when there is a rectangle to be printed */ for(j=col1; j<=col2; j++) printf("%d ", a[i][j]); for(j=col2,i=row1+1; i<=row2; i++) printf("%d ", a[i][j]); for(i=row2,j=col2-1; j>=col1; j--) printf("%d ", a[i][j]); for(j=col1,i=row2-1; i>row1; i--) printf("%d ", a[i][j]); } 

Answer 17

Here is my implementation in Java:

 public class SpiralPrint { static void spiral(int a[][],int x,int y){ //If the x and y co-ordinate collide, break off from the function if(x==y) return; int i; //Top-left to top-right for(i=x;i<y;i++) System.out.println(a[x][i]); //Top-right to bottom-right for(i=x+1;i<y;i++) System.out.println(a[i][y-1]); //Bottom-right to bottom-left for(i=y-2;i>=x;i--) System.out.println(a[y-1][i]); //Bottom left to top-left for(i=y-2;i>x;i--) System.out.println(a[i][x]); //Recursively call spiral spiral(a,x+1,y-1); } public static void main(String[] args) { int a[][]={{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}}; spiral(a,0,4); /*Might be implemented without the 0 on an afterthought, all arrays will start at 0 anyways. The second parameter will be the dimension of the array*/ } } 

Answer 18

 //shivi..coding is adictive!! #include<shiviheaders.h> #define R 3 #define C 6 using namespace std; void PrintSpiral(int er,int ec,int arr[R][C]) { int sr=0,sc=0,i=0; while(sr<=er && sc<=ec) { for(int i=sc;i<=ec;++i) cout<<arr[sr][i]<<" "; ++sr; for(int i=sr;i<=er;++i) cout<<arr[i][ec]<<" "; ec--; if(sr<=er) { for(int i=ec;i>=sc;--i) cout<<arr[er][i]<<" "; er--; } if(sc<=ec) { for(int i=er;i>=sr;--i) cout<<arr[i][sc]<<" "; ++sc; } } } int main() { int a[R][C] = { {1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18} }; PrintSpiral(R-1, C-1, a); } 

Answer 19

Java code if anyone is interested.
Input :
four
1 2 3 4
5 6 7 8
9 1 2 3
4 5 6 7

Output : 1 2 3 4 8 3 7 6 5 4 9 5 6 7 2 1

 public class ArraySpiralPrinter { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); //marrix size //read array int[][] ar = new int[n][n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { ar[i][j] = sc.nextInt(); } } printTopRight(0, 0, n - 1, n - 1, ar); } //prints top and right layers. //(x1,y1) to (x1, y2) - top layer & (x1,y2) to (x2, y2) private static void printTopRight(int x1, int y1, int x2, int y2, int[][] ar) { //print row values - top for (int y = y1; y <= y2; y++) { System.out.printf("%d ", ar[x1][y]); } //print column value - right for (int x = x1 + 1; x <= x2; x++) { System.out.printf("%d ", ar[x][y2]); } //are there any remaining layers if (x2 - x1 > 0) { //call printBottemLeft printBottomLeft(x1 + 1, y1, x2, y2 - 1, ar); } } //prints bottom and left layers in reverse order //(x2,y2) to (x2, y1) - bottom layer & (x2,y1) to (x1, y1) private static void printBottomLeft(int x1, int y1, int x2, int y2, int[][] ar) { //print row values in reverse order - bottom for (int y = y2; y >= y1; y--) { System.out.printf("%d ", ar[x2][y]); } //print column value in reverse order - left for (int x = x2-1; x >= x1; x--) { System.out.printf("%d ", ar[x][y1]); } //are there any remaining layers if (x2 - x1 > 0) { printTopRight(x1, y1 + 1, x2 - 1, y2, ar); } } } 

Answer 20

This is a recursive version in C that I could think of: -

 void printspiral (int[][100],int, int, int, int); int main() { int r,c, i, j; printf ("Enter the dimensions of the matrix"); scanf("%d %d", &r, &c); int arr[r][100]; int min = (r<c?r:c); if (min%2 != 0) min = min/2 +1; for (i = 0;i<r; i++) for (j = 0; j<c; j++) scanf ("%d",&arr[i][j]); printspiral(arr,0,r,c,min ); } void printspiral (int arr[][100], int i, int j, int k, int min) { int a; for (a = i; a<k;a++) printf("%d\n", arr[i][a]); for (a=i+1;a<j;a++) printf ("%d\n", arr[a][k-1]); for (a=k-2; a>i-1;a--) printf("%d\n", arr[j-1][a]); for (a=j-2; a>i; a--) printf("%d\n", arr[a][i]); if (i < min) printspiral(arr,i+1, j-1,k-1, min); } 

Answer 21

http://www.technicalinterviewquestions.net/2009/03/print-2d-array-matrix-spiral-order.html

here is the best explanation of the answer above :) along with the chart :)

Answer 22

 public static void printSpiral1(int array[][],int row,int col){ int rowStart=0,colStart=0,rowEnd=row-1,colEnd=col-1; int i; while(rowStart<=rowEnd && colStart<= colEnd){ for(i=colStart;i<=colEnd;i++) System.out.print(" "+array[rowStart][i]); for(i=rowStart+1;i<=rowEnd;i++) System.out.print(" "+array[i][colEnd]); for(i=colEnd-1;i>=colStart;i--) System.out.print(" "+array[rowEnd][i]); for(i=rowEnd-1;i>=rowStart+1;i--) System.out.print(" "+array[i][colStart]); rowStart++; colStart++; rowEnd--; colEnd--; } } 

Answer 23

 public class SpiralPrint{ //print the elements of matrix in the spiral order. //my idea is to use recursive, for each outer loop public static void printSpiral(int[][] mat, int layer){ int up = layer; int buttom = mat.length - layer - 1; int left = layer; int right = mat[0].length - layer - 1; if(up > buttom+1 || left > right + 1) return; // termination condition //traverse the other frame, //print up for(int i = left; i <= right; i ++){ System.out.print( mat[up][i]+ " " ); } //print right for(int i = up + 1; i <=buttom; i ++){ System.out.print(mat[i][right] + " "); } //print buttom for(int i = right - 1; i >= left; i --){ System.out.print(mat[buttom][i] + " "); } //print left for(int i = buttom - 1; i > up; i --){ System.out.print(mat[i][left] + " "); } //recursive call for the next level printSpiral(mat, layer + 1); } public static void main(String[] args){ int[][] mat = {{1,2,3,4}, {5,6,7,8}, {9,10,11,12}, {13,14,15,16}}; int[][] mat2 = {{1,2,3}, {4,5,6}, {7,8,9}, {10,11,12}}; SpiralPrint.printSpiral(mat2,0); return; } } 

Answer 24

Here is my solution in C #:

 public static void PrintSpiral(int[][] matrix, int n) { if (matrix == null) { return; } for (int layer = 0; layer < Math.Ceiling(n / 2.0); layer++) { var start = layer; var end = n - layer - 1; var offset = end - 1; Console.Write("Layer " + layer + ": "); // Center case if (start == end) { Console.Write(matrix[start][start]); } // Top for (int i = start; i <= offset; i++) { Console.Write(matrix[start][i] + " "); } // Right for (int i = start; i <= offset; i++) { Console.Write(matrix[i][end] + " "); } // Bottom for (int i = end; i > start; i--) { Console.Write(matrix[end][i] + " "); } // Left for (int i = end; i > start; i--) { Console.Write(matrix[i][start] + " "); } Console.WriteLine(); } } 

Answer 25

Here is my approach using Iterator. Please note that this solves almost the same problem. Full code here: https://github.com/rdsr/algorithms/blob/master/src/jvm/misc/FillMatrix.java

 import java.util.Iterator; class Pair { final int i; final int j; Pair(int i, int j) { this.i = i; this.j = j; } @Override public String toString() { return "Pair [i=" + i + ", j=" + j + "]"; } } enum Direction { N, E, S, W; } class SpiralIterator implements Iterator<Pair> { private final int r, c; int ri, ci; int cnt; Direction d; // current direction int level; // spiral level; public SpiralIterator(int r, int c) { this.r = r; this.c = c; d = Direction.E; level = 1; } @Override public boolean hasNext() { return cnt < r * c; } @Override public Pair next() { final Pair p = new Pair(ri, ci); switch (d) { case E: if (ci == c - level) { ri += 1; d = changeDirection(d); } else { ci += 1; } break; case S: if (ri == r - level) { ci -= 1; d = changeDirection(d); } else { ri += 1; } break; case W: if (ci == level - 1) { ri -= 1; d = changeDirection(d); } else { ci -= 1; } break; case N: if (ri == level) { ci += 1; level += 1; d = changeDirection(d); } else { ri -= 1; } break; } cnt += 1; return p; } private static Direction changeDirection(Direction d) { switch (d) { case E: return Direction.S; case S: return Direction.W; case W: return Direction.N; case N: return Direction.E; default: throw new IllegalStateException(); } } @Override public void remove() { throw new UnsupportedOperationException(); } } public class FillMatrix { static int[][] fill(int r, int c) { final int[][] m = new int[r][c]; int i = 1; final Iterator<Pair> iter = new SpiralIterator(r, c); while (iter.hasNext()) { final Pair p = iter.next(); m[pi][pj] = i; i += 1; } return m; } public static void main(String[] args) { final int r = 19, c = 19; final int[][] m = FillMatrix.fill(r, c); for (int i = 0; i < r; i++) { for (int j = 0; j < c; j++) { System.out.print(m[i][j] + " "); } System.out.println(); } } } 

Answer 26

End a clean C program for any matrix in a 2D array with a given column of row x.

 #include <stdio.h> void printspiral(int *p,int r, int c) { int i=0,j=0,m=1,n=0; static int firstrun=1,gCol; if (!p||r<=0||c<=0) return ; if(firstrun) { gCol=c; firstrun=0; } for(i=0,j=0;(0<=i && i<c)&&(0<=j && j<r);i+=m,j+=n) { printf(" %d",p[i+j*gCol]); if (i==0 && j==1 && (i+1)!=c) break; else if (i+1==c && !j) {m=0;n=1;} else if (i+1==c && j+1==r && j) {n=0;m=-1;} else if (i==0 && j+1==r && j) {m=0;n=-1;} } printspiral(&p[i+j*gCol+1],r-2,c-2); firstrun=1; printf("\n"); } int main() { int a[3][3]={{0,1,2},{3,4,5},{6,7,8}}; int b[3][4]={{0,1,2,3},{4,5,6,7},{8,9,10,11}}; int c[4][3]={{0,1,2},{3,4,5},{6,7,8},{9,10,11}}; int d[3][1]={{0},{1},{2}}; int e[1][3]={{0,1,2}}; int f[1][1]={{0}}; int g[5][5]={{0,1,2,3,4},{5,6,7,8,9},{10,11,12,13,14},{15,16,17,18,19},{20,21,22,23,24}}; printspiral(a,3,3); printspiral(b,3,4); printspiral(c,4,3); printspiral(d,3,1); printspiral(e,1,3); printspiral(f,1,1); printspiral(g,5,5); return 0; } 

Answer 27

: php

, , , . - , , , , . PHP :

 #The source number matrix $source[0] = array(1, 2, 3, 4); $source[1] = array(5, 6, 7, 8); $source[2] = array(9, 10, 11, 12); $source[3] = array(13, 14, 15, 16); $source[4] = array(17, 18, 19, 20); #Get the spiralled numbers $final_spiral_list = get_spiral_form($source); print_r($final_spiral_list); function get_spiral_form($matrix) { #Array to hold the final number list $spiralList = array(); $result = $matrix; while(count($result) > 0) { $resultsFromRead = get_next_number_circle($result, $spiralList); $result = $resultsFromRead['new_source']; $spiralList = $resultsFromRead['read_list']; } return $spiralList; } function get_next_number_circle($matrix, $read) { $unreadMatrix = $matrix; $rowNumber = count($matrix); $colNumber = count($matrix[0]); #Check if the array has one row or column if($rowNumber == 1) $read = array_merge($read, $matrix[0]); if($colNumber == 1) for($i=0; $i<$rowNumber; $i++) array_push($read, $matrix[$i][0]); #Check if array has 2 rows or columns if($rowNumber == 2 || ($rowNumber == 2 && $colNumber == 2)) { $read = array_merge($read, $matrix[0], array_reverse($matrix[1])); } if($colNumber == 2 && $rowNumber != 2) { #First read left to right for the first row $read = array_merge($read, $matrix[0]); #Then read down on right column for($i=1; $i<$rowNumber; $i++) array_push($read, $matrix[$i][1]); #..and up on left column for($i=($rowNumber-1); $i>0; $i--) array_push($read, $matrix[$i][0]); } #If more than 2 rows or columns, pick up all the edge values by spiraling around the matrix if($rowNumber > 2 && $colNumber > 2) { #Move left to right for($i=0; $i<$colNumber; $i++) array_push($read, $matrix[0][$i]); #Move top to bottom for($i=1; $i<$rowNumber; $i++) array_push($read, $matrix[$i][$colNumber-1]); #Move right to left for($i=($colNumber-2); $i>-1; $i--) array_push($read, $matrix[$rowNumber-1][$i]); #Move bottom to top for($i=($rowNumber-2); $i>0; $i--) array_push($read, $matrix[$i][0]); } #Now remove these edge read values to create a new reduced matrix for the next read $unreadMatrix = remove_top_row($unreadMatrix); $unreadMatrix = remove_right_column($unreadMatrix); $unreadMatrix = remove_bottom_row($unreadMatrix); $unreadMatrix = remove_left_column($unreadMatrix); return array('new_source'=>$unreadMatrix, 'read_list'=>$read); } function remove_top_row($matrix) { $removedRow = array_shift($matrix); return $matrix; } function remove_right_column($matrix) { $neededCols = count($matrix[0]) - 1; $finalMatrix = array(); for($i=0; $i<count($matrix); $i++) $finalMatrix[$i] = array_slice($matrix[$i], 0, $neededCols); return $finalMatrix; } function remove_bottom_row($matrix) { unset($matrix[count($matrix)-1]); return $matrix; } function remove_left_column($matrix) { $neededCols = count($matrix[0]) - 1; $finalMatrix = array(); for($i=0; $i<count($matrix); $i++) $finalMatrix[$i] = array_slice($matrix[$i], 1, $neededCols); return $finalMatrix; } 

Answer 28

 // Program to print a matrix in spiral order #include <stdio.h> int main(void) { // your code goes here int m,n,i,j,k=1,c1,c2,r1,r2;; scanf("%d %d",&m,&n); int a[m][n]; for(i=0;i<m;i++) { for(j=0;j<n;j++) { scanf("%d",&a[i][j]); } } r1=0; r2=m-1; c1=0; c2=n-1; while(k<=m*n) { for(i=c1;i<=c2;i++) { k++; printf("%d ",a[r1][i]); } for(j=r1+1;j<=r2;j++) { k++; printf("%d ",a[j][c2]); } for(i=c2-1;i>=c1;i--) { k++; printf("%d ",a[r2][i]); } for(j=r2-1;j>=r1+1;j--) { k++; printf("%d ",a[j][c1]); } c1++; c2--; r1++; r2--; } return 0; } 

Answer 29

java- mx n. = =

 public static void printSpiral(int rows, int columns, int a[][]) { int i, k = 0, l = 0; /* k - starting row index l - starting column index */ while (k < rows && l < columns) { /* Print the first row from the remaining rows */ for (i = l; i < columns; ++i) { System.out.println(a[k][i]); } k++; /* Print the last column from the remaining columns */ for (i = k; i < rows; ++i) { System.out.println(a[i][columns-1]); } columns--; /* Print the last row from the remaining rows */ if ( k < rows) { for (i = columns-1; i >= l; --i) { System.out.println(a[rows-1][i]); } rows--; } /* Print the first column from the remaining columns */ if (l < columns) { for (i = rows-1; i >= k; --i) { System.out.println(a[i][l]); } l++; } } } 

Answer 30

Here is my solution. , , .

 class Spiral: def spiralOrder(self, A): result = [] c = [] c.append(A[0]) b = A[1:] while len(b) > 0: b = self.rotate(b) c.append(b[0]) b = b[1:] for item in c: for fitem in item: print fitem, result.append(fitem) return result def rotate(self,a): b = [] l = zip(*a) for i in xrange(len(l)-1,-1,-1): b.append(list(l[i])) return b if __name__ == '__main__': a = [[1, 2, 3,3], [4, 5, 6,6], [7, 8, 9,10]] s = Spiral() s.spiralOrder(a)