java - Reverse words of a char[]/sentence in place -

i want reverse words of string in place using constant space. catch words cannot contain consecutive whitespaces. want reduce consecutive whitespaces between words 1 whitespace , ignore leading , trailing whitespaces. able implement in place reversal of words, i'm struggling implement removing consecutive whitespaces between words , leading , trailing whitespaces. can me?

here have far:

public char[] reversewords(char[] s) {      if (s == null)         return null;      int right = 0;       s = reversestring(s, 0, s.length - 1);// reversed sentence     //system.out.println(s);     (int left = 0; left < s.length; left++) {          if (s[left] != ' ') {// first word              (right = left; right < s.length && s[right] != ' '; right++)                 ; // end of word              s = reversestring(s, left, right - 1);             left =(right - 1);// move left index end of                                                 // word              // s[left++] = ' ';         }      }     return s; }  public char[] reversestring(char[] strchars, int start, int end) {      if (strchars == null)         return null;      while (start < end) {          char temp = strchars[start];         strchars[start] = strchars[end];         strchars[end] = temp;          start++;         end--;     }     return strchars; } 

there going easier/faster ways. giving thought should in learning purpose.

first, whatever have now, reverse words, , leave consecutive space untouched.

then write method consecutive space removing.

have 2 pointer, start @ first position not space.

a , b keep on moving on together.

if (a != b), s[a] = s[b]; s[b] = ' ';

if s[a] , s[a-1] space, stop (a @ 2nd space), , b continue moving forward. such way, stay same position , continue copy b, until b gives non-space character.

and ends when b hit end.

in psuedo code, like

int = first position of non-space; int b = a;  while b < s.size() {   if (a != b) {     s[a] = s[b]     s[b] = ' '   }   if (both s[a] , s[a-1] space)  {      increment b;     // leave untouched   } else {     increment a;     increment b;   } } 

constant space, o(n) time

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another way, can handle removal of consecutive space in place when reversing words:

the hint is, include spaces when doing reverse.

e.g. given string

abc    def     ghi l                 (left)     

first reverse trivial skipping it. hint is, second word, going stop l @ position right after first space:

cba    def     ghi     l 

the "right" side of reverse going first right boundary of word:

cba    def     ghi     l    r 

then reverse in place

cba fed        ghi     l    r 

then keep on finding next position of l start reverse again:

cba fed        ghi         l 

take similar logic

cba fed        ghi         l        r 

then

cba fed ihg                l        r