Word inverse string word in Word

Question

I have the following code to cancel line by word, I have a question, although at first one could indicate how to make the code better? and secondly, how can I remove the space in which I end at the beginning of a new line.

String str = "hello brave new world"; tStr.reverseWordByWord(str) public String reverseWordByWord(String str){ int strLeng = str.length()-1; String reverse = "", temp = ""; for(int i = 0; i <= strLeng; i++){ temp += str.charAt(i); if((str.charAt(i) == ' ') || (i == strLeng)){ for(int j = temp.length()-1; j >= 0; j--){ reverse += temp.charAt(j); if((j == 0) && (i != strLeng)) reverse += " "; } temp = ""; } } return reverse; } 

the phrase is currently becoming:

olleh evarb wen dlrow

note the space at the beginning of a new line.

Answer 1

Without using the split function, the code will look like this:

 public static void reverseSentance(String str) { StringBuilder revStr = new StringBuilder(""); int end = str.length(); // substring takes the end index -1 int counter = str.length()-1; for (int i = str.length()-1; i >= 0; i--) { if (str.charAt(i) == ' ' || i == 0) { if (i != 0) { revStr.append(str.substring(i+1, end)); revStr.append(" "); } else { revStr.append(str.substring(i,end)); } end = counter; } counter--; } System.out.println(revStr); } 

Answer 2

If str = "The quick brown fox jumped over the lazy dog!" he will bring it back like "dog! lazy over the brown fox jumping fast" ...

  private static String Reverse(String str) { char charArray[] = str.toCharArray(); for (int i = 0; i <str.length(); i++){ if(charArray[i] == ' ') return Reverse(str.substring(i + 1)) + str.substring(0, i) + " "; } return str + " "; } 

Answer 3

Here is how you could do it:

  StringBuilder result = new StringBuilder(); StringTokenizer st = new StringTokenizer(input, " "); while (st.hasMoreTokens()) { StringBuilder thisToken = new StringBuilder(st.nextToken()); result.append(thisToken.reverse() + " "); } String resultString = result.toString(); 

Answer 4

My approach using StringUtils. In unit test.

 @Test public void testReversesWordsAndThenAllCharacters(){ String sentence = "hello brave new world"; String reversedWords = StringUtils.reverseDelimited(sentence, ' '); String reversedCharacters = StringUtils.reverse(reversedWords); assertEquals("olleh evarb wen dlrow", reversedCharacters); } 

If you are importing StringUtils statically, this might be included:

 reverse(reverseDelimited("hello brave new world", ' ')) 

Answer 5

The first thing I would like to do is split the code that modifies the words with the code that modifies each word individually. This inner loop:

 for(int j = temp.length()-1; j >= 0; j--) { reverse += temp.charAt(j); if((j == 0) && (i != strLeng)) reverse += " "; } 

There will be a function / method call.

Also, to make your code more efficient, instead of concatenating strings with the + operator, I would use a string buffer class. For example StringBuffer or StringBuilder .

Answer 6

First of all, you should divide it into three functions. The first splitting of a large line in the list of lines using space as a separator, the second change of one line without spaces and the last lines of concatenation.

When you do this, it will be easier for you to find the cause of the space. You can already see what's in the current code, but I'm not going to tell you: D.

Answer 7

How about using something like that?

 String string="yourWord"; String reverse = new StringBuffer(string).reverse().toString(); 

Answer 8

You can use StringUtils

 return StringUtils.reverseDelimitedString(str, " "); 

Answer 9

Try it. It takes into account punctuation and whitespace characters of any type.

 public String reverseWordByWord(String inputStr) { BreakIterator wordIterator = BreakIterator.getWordInstance(); wordIterator.setText(inputStr); int start = wordIterator.first(); StringBuilder tempBuilder; StringBuilder outBuilder = new StringBuilder(); for (int end = wordIterator.next(); end != BreakIterator.DONE; start = end, end = wordIterator.next()) { tempBuilder = new StringBuilder(inputStr.substring(start, end)); outBuilder.append(tempBuilder.reverse()); } return outBuilder.toString(); } 

Answer 10

 public class StringReversers { public static void main(String[] args) { String s = new String(revStr("hello brave new world")); String st = new String(revWords("hello brave new world")); System.out.println(s); System.out.println(st); } public static String revStr(String s){ StringBuilder sb = new StringBuilder(); for (int i=s.length()-1; i>=0;i--){ sb.append(s.charAt(i)); } return sb.toString(); } public static String revWords(String str) { StringBuilder sb = new StringBuilder(); String revd = revStr(str); for (String s : revd.split(" ")){ sb.append(revStr(s)); sb.append(" "); } return sb.toString(); } } 

Answer 11

 public static void reverseByWord(String s) { StringTokenizer token = new StringTokenizer(s); System.out.println(token.countTokens()); Stack<String> stack = new Stack<String>(); while (token.hasMoreElements()) { stack.push(token.nextElement().toString()); } while (!stack.isEmpty()) { System.out.println(stack.pop()); } } 

Answer 12

Another solution without using the split method

  public static String reverseWordsWithoutSplit(String str) { StringBuffer buffer = new StringBuffer(); int length = str.length(); while(length >0) { int wordstart = length -1; while(wordstart >0 && str.charAt(wordstart) != ' '){ wordstart--; } buffer.append(str.substring(wordstart==0?wordstart:wordstart+1, length)); if(wordstart>0) buffer.append(" "); length = wordstart; } return buffer.toString(); } 

Answer 13

Another solution. This decision is in place.

Inverse words in a string (words are separated by one or more spaces), spaces can precede a word (s), i.e. a space at the beginning of the sentence, the end, etc .... Solve it in place

 public class ReverseWordsInString { public static void main(String[] args) { // TODO Auto-generated method stub char[] sentence = " Hi my name is person!".toCharArray(); System.out.println(ReverseSentence(sentence)); } private static char[] ReverseSentence(char[] sentence) { //Given: "Hi my name is person!" //produce: "iH ym eman si !nosrep" //the obvious naive solution: utilize stringtokenize to separate each word into its own array. reverse each word and insert space between each array print //better solution: drop stringtokenize and use a counter to count how many characters processed before space was hit. // once space hit, then jump back swap characters between counter-1 and start position. O(1) Space if(sentence == null) return null; if(sentence.length == 1) return sentence; int startPosition=0; int counter = 0; int sentenceLength = sentence.length-1; //Solution handles any amount of spaces before, between words etc... while(counter <= sentenceLength) { if(sentence[counter] == ' ' && startPosition != -1 || sentenceLength == counter) //Have passed over a word so upon encountering a space or end of string reverse word { //swap from startPos to counter - 1 //set start position to -1 and increment counter int begin = startPosition; int end; if(sentenceLength == counter) { end = counter; } else end = counter -1; char tmp; //Reverse characters while(end >= begin){ tmp = sentence[begin]; sentence[begin] = sentence[end]; sentence[end] = tmp; end--; begin++; } startPosition = -1; //flag used to indicate we have no encountered a character of a string } else if(sentence[counter] !=' ' && startPosition == -1) //first time you encounter a letter in a word set the start position { startPosition = counter; } counter++; } return sentence; } 

}

Answer 14

Here, one thread also discusses this issue. I think the one answer using split with regex is very smart.

https://codereview.stackexchange.com/questions/43838/reverse-a-string-word-by-word

 public String reverseWordByWord(String s) { StringBuilder result = new StringBuilder(); String[] words = sentence.split("\\s+"); for (int i = words.length - 1 ; 0 <= i; i--) { result.append(words[i]).append(' '); } return result.toString().trim(); } 

Answer 15

The answer to removing the leading space character is easy, simple

 return reverse.trim(); 

String.trim() returns a copy of the string with the missing leading and trailing spaces (as copied from the Javadoc documentation).

For your common problem, I made this example:

 String job = "This is a job interview question!"; StringBuilder sb = new StringBuilder(job); String[] words = job.split(" "); int i = 0; for (String word : words) { words[i] = (new StringBuilder(word)).reverse().toString(); i++; } System.out.println("job = " + job); System.out.print("rev = "); for (String word: words) { sb.append(new StringBuilder(word).toString()); sb.append(" "); } String rev = sb.toString().trim(); System.out.println(rev); 

and output:

 job = This is a job interview question! rev = sihT si a boj weivretni !noitseuq 

If you want to include any space character more, for example, a tab character, line break, form feed, change the split() argument to split("\\s") , since \s is the regular expression character class that represents [\ t \ r \ n \ F]. Note how you should avoid the backslash character in the Java string representation of the regular expression (this is what the split method expects).

Answer 16

how to undo a word in java

 public class ReverseString { public static void main(String[] args) { String reverse = ""; String original = new String("hidaya"); for ( int i = original.length() - 1 ; i >= 0 ; i-- ) reverse = reverse + original.charAt(i); System.err.println("Orignal string is: "+original); System.out.println("Reverse string is: "+reverse); } } 

Answer 17

Here is a coding technique using the popular split () function, which is available in all major Java languages toCharArray (), good for full character control in string form, and the Java StringBuilder class for performance (available in C # too).

I think the code is easier to understand compared to other posted answers

 public static String reverseWordByWord(String sentence) { StringBuilder result = new StringBuilder(); String[] words = sentence.split("\\s+"); // space(s) are the delimiters for (String word : words) { char[] charArray = word.toCharArray(); int iEnd = word.length() - 1; StringBuilder temp = new StringBuilder(); for (int i = iEnd; i >= 0; i--) { temp.append(charArray[ i]); } result.append(temp); result.append(" "); // separate the words } return result.toString().trim(); // remove the trailing spaces } 

Reminder of copyright claims.
Input Example: "Hello World"
Exit : "olleH dlroW"

Answer 18

The following should do it in O (n) without any costly copying the array or restructuring the length of the character array. Consideration of several preceding, intermediate and trailing spaces.

 public class ReverseString { public static void main(String[] args) { String string1 = "hello brave new world"; String string2 = "hello brave new world "; String string3 = " hello brave new world"; String string4 = " hello brave new world "; System.out.println(reverseStringWordByWord(string1)); System.out.println(reverseStringWordByWord(string2)); System.out.println(reverseStringWordByWord(string3)); System.out.println(reverseStringWordByWord(string4)); } private static String reverseStringWordByWord(String string) { StringBuilder sb = new StringBuilder(); int length = string.length(); for(int i=0;i<length;i++) { char c = string.charAt(i); if(c == ' ') { sb.append(c); } else { int j = i; while(j < length && string.charAt(j) != ' ') { j++; } sb.append(reverseString(string.substring(i, j))); i = j-1; } } return sb.toString(); } private static String reverseString(String string) { StringBuilder sb = new StringBuilder(); for(int i=string.length()-1;i>=0; i--) { sb.append(string.charAt(i)); } return sb.toString(); } } 

Answer 19

 public String reverseStringWordByWord(String input) { StringBuilder returnValue = new StringBuilder(); int insertIndex = 0; for(int i = 0;i < input.length();i++ ) { if(input.charAt(i)!=' ') { returnValue.insert(insertIndex, currentChar); } else { insertIndex = i+1; returnValue.append(currentChar); } } return returnValue.toString(); } 

Answer 20

I am new to Java myself and I expect to be beaten before the hit, but I thought I would give it to him anyway. You can solve the problem with an extra space by creating a line with the assumption that you will remove the unnecessary extra space at the end. If performance is a consideration, then you can rethink this!

Edit: Please note that my solution (now) handles leading and trailing spaces.

 public class StringReversal { public static void main(String[] args) { String str = "hello brave new world"; System.out.println("\"" + reverseWordByWord(str) + "\""); } public static String reverseWordByWord(String str) { String reverse = ""; boolean first = true; for (String s : str.split(" ")) { if (first) { first = false; } else { reverse += " "; } StringBuilder sb = new StringBuilder(); for (int i = s.length() - 1; i >= 0; --i) { sb.append(s.charAt(i)); } reverse += sb.toString(); } while (reverse.length() < str.length()) { reverse += " "; } return reverse.substring(0, reverse.length()); } } 

Answer 21

So, I assume that you are learning / practicing java and that there is a high risk of the homework question ... This means that you will love or hate this answer ...

If you look at the source code of a String object, you will find something like this inside:

private final char value[]; //this stores the String characters

The first step is to get this value [] with:

 char[] myChars = str.toCharArray(); 

Pay attention to the implementation of the function (from openjdk-7 ), it returns a copy of the array, not the original one, since String objects are immutable.

 public char[] toCharArray() { char result[] = new char[count]; getChars(0, count, result, 0); //Calls System.arraycopy(...) return result; } 

Now that we have myChars , we can play with it and get the result in linear time O (n)!

 public static String reverseWordByWord(String str) { char[] myChars = str.toCharArray(); int stringLen = myChars.length; int left = 0, right = 0; for(int index = 0; index < stringLen; index++) { if(chars[index] == ' ') { //assign right reverse(chars, left, right); //update left } } //Don't forget to handle the boundary case (last word in the String)! } 

And here is the inverse function:

 private static void reverse(char[] chars, int left, int right) { while(left < right) { //Would you know how to swap 2 chars without using a "char tmp" variable? ;) //Update left and right } } 

Now, just for fun, you might want to try to get the following result, and perhaps you will get this exact question from some interviewer who once completed a fantasy:

world new brave hello

Answer 22

Take a string and using the Stack methods with the StringTokenizer Object and with our methods we can cut the string into a piece of words using delimeter . Using the Stack Natural function, paste (push) all words in Satck and Remove (pop) all words from Stack. then print everything.

Here we can take String s = "hello brave new world"

 import java.util.*; public class StringReverse { public static void main(String[] argv) { String s = "hello brave new world"; Stack<String> myStack = new Stack<String>(); StringTokenizer st = new StringTokenizer(s); while (st.hasMoreTokens()) myStack.push((String) st.nextElement()); // Print the stack backwards System.out.print('"' + s + '"' + " backwards by word is:\n\t\""); while (!myStack.empty()) { System.out.print(myStack.pop()); System.out.print(' '); } System.out.println('"'); } } 

If you are using any package with your own, then check the output of the above program .

Answer 23

  StringBuilder sb = " This is cool"; sb.reverse(); //sb now contains "looc si sihT " System.out.println(sb); for(int i = 0; i < sb.length(); i++) { int index = sb.indexOf(" ", i); // System.out.println(index); if(index > 0) { sb.replace(i, index, new StringBuilder(sb.substring(i, index)).reverse().toString()); i = index; } if(index < 0) { sb.replace(i, sb.length(), new StringBuilder(sb.substring(i, sb.length())).reverse().toString()); break; } } System.out.println(sb); //output "cool is This " 

Answer 24

  // Create Scanner object Scanner s=new Scanner(System.in); // Take no.of strings that the user wants int n=s.nextInt(); // Create a temp array String temps[]=new String[n]; // Initialize the variable before the user input is stored in it String st=""; // Create a words array String words[]; // Skip first line, if not used user input will be skipped one time s.nextLine(); // Read the no.of strings that user wish to.. for(int k=0;k<n;k++) { System.out.println("String #"+(k+1)+": "); // Read line st=s.nextLine(); // Split words with a space, because words has spaces at start, end positions. words=st.split(" "); // Initialize temps[k] to avoid null temps[k]=""; // Reverse string now! for(int i=words.length-1;i>=0;i--) { // Put each word in user input string from end to start with a space temps[k]+=words[i]+" "; } } // Now print the words! for(int i=0;i<n;i++) { // Print the reversed strings, trim them to avoid space at last, see in the reverse logic loop, space is attached at last! System.out.println("String #"+(i+1)+": "+temps[i].trim()); } 

Answer 25

I think my code below is more efficient than any code available here:

 public static void revWordsInStringCStyle(String str){ char [] str_ch = str.toCharArray(); System.out.println(str); char temp; int len = str_ch.length; int left = len-1; for(int right =0; right<len/2 ;right++){ temp = str_ch[left]; str_ch[left] = str_ch[right]; str_ch[right] = temp; left--; } for(int i =0; i < len ; i++){ System.out.print(str_ch[i]); } 

}

Example: "hello world"

will become: "dlrow olleho"

Answer 26

How about this:

  public class Main { public static void main(String args[]){ String input ="***NGuyen**Van******A*******"; String temp = ""; String result =""; for( int i = 0 ; i < input.length() ; i++) { if(input.charAt(i) != '*') { temp = temp + input.charAt(i); } else { if(!temp.equals("")) result = temp + result; result = input.charAt(i) + result ; temp =""; } } System.out.println(result); } } Output: *******A******Van**NGuyen*** 

Answer 27

I tried to do without the split function. Use a substring for a loop instead.

 static String reverseSentenceWithoutSplit(String str){ StringBuilder sb = new StringBuilder(); char [] charArray = str.toCharArray(); int endindex = charArray.length-1; // loop in reverse, char by char for(int i=charArray.length-1; i>=0; i--){ char c = charArray[i]; if(c==' '){ sb.append(str.substring(i + 1, endindex+1)); // substring- start index inclusive, end index exclusive endindex=i-1;// move to first letter sb.append(c); // include the space } if(i==0){ //grab the last word sb.append(str.substring(i, endindex+1)); } } if(sb.length()==0){ // handle case where string has no space return str; } return sb.toString(); } 

Entrance: you are the symbol of oppression; Output: oppression of the symbol a

Input: ThisIsAllOneWord Output: ThisIsAllOneWord

Answer 28

 public String reverseEach(String input) { String[] test = input.split(" "); String output=""; for(String t:test) { String p =""; for(int i=t.length()-1;i>=0;i--) { p=p+t.charAt(i); } output=output+p+" "; } return output; } 

Answer 29

 /* this code uses while loop and the position of spaces come correctly which is a problem if you use for loop */ import java.util.*; class StrWordRev { public void rev(String s) { for(int i=s.length()-1;i>=0;i--) { System.out.print(s.charAt(i)); } System.out.print(" "); } public void main() { Scanner sc=new Scanner(System.in); String s,s1=""; System.out.println("Enter the string : "); s=sc.nextLine(); int i=0; while(i<s.length()) { s1=""; while(i<s.length() && s.charAt(i)!=' ') { s1=s1+s.charAt(i); i++; } rev(s1); i=i+1; } } 

Answer 30

This should work for you:

 import java.io.*; class internal1 { public static void main(String s[] { DataInputStream dis = new DataInputStream(System.in); try { String a = ""; String b = ""; System.out.print("Enter the string::"); a = dis.readLine(); System.out.print(a.length()); System.out.println(" "); for (int i = 0; i <= a.length() - 1; i++) { if (a.charAt(i) == ' ' || a.charAt(i) == '.') { for (int j = b.length() - 1; j >= 0; j--) { System.out.print(b.charAt(j)); } b = ""; System.out.print(" "); } b = b + a.charAt(i); } } catch (Exception e) { } } }