leetcode186. 翻转字符串里的单词 II

给定一个字符串，逐个翻转字符串中的每个单词。

示例：

输入: ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
输出: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]
注意：

单词的定义是不包含空格的一系列字符
输入字符串中不会包含前置或尾随的空格
单词与单词之间永远是以单个空格隔开的
进阶：使用 O(1) 额外空间复杂度的原地解法。

思路：先反转每个单词，然后总体再翻转。

  

   

    

     

    
    

     

      class Solution {
     
    
   

    

     

    
    

     
 public void reverseWords(char[] s) {
     
    
   

    

     

    
    

     
 int start=0;
     
    
   

    

     

    
    

     
 for(int i=0;i<s.length;i++){
     
    
   

    

     

    
    

     
 if(s[i]==' '){
     
    
   

    

     

    
    

     

       reverseWord(s,start,i-1);
     
    
   

    

     

    
    

     

       start=i+1;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       reverseWord(s,start,s.length-1);
     
    
   

    

     

    
    

     

       reverseWord(s,0,s.length-1);
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     
 public void reverseWord(char[] s,int start,int end){
     
    
   

    

     

    
    

     
 char temp;
     
    
   

    

     

    
    

     
 while(start<end){
     
    
   

    

     

    
    

     

       temp=s[start];
     
    
   

    

     

    
    

     

       s[start]=s[end];
     
    
   

    

     

    
    

     

       s[end]=temp;
     
    
   

    

     

    
    

     

       start++;
     
    
   

    

     

    
    

     

       end--;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: fantianzuo.blog.csdn.net，作者：兔老大RabbitMQ，版权归原作者所有，如需转载，请联系作者。

原文链接：fantianzuo.blog.csdn.net/article/details/104100599