Given an input string , reverse the string word by word. 

Example:

Input:  ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

Note: 

• A word is defined as a sequence of non-space characters.
• The input string does not contain leading or trailing spaces.
• The words are always separated by a single space.

Follow up: Could you do it in-place without allocating extra space?

 

这道题让我们翻转一个字符串中的单词，跟之前那题  没有区别，由于之前那道题我们就是用in-place的方法做的，而这道题反而更简化了题目，因为不考虑首尾空格了和单词之间的多空格了，方法还是很简单，先把每个单词翻转一遍，再把整个字符串翻转一遍，或者也可以调换个顺序，先翻转整个字符串，再翻转每个单词，参见代码如下：

 

解法一：

class Solution {public:    void reverseWords(vector
     
      & str) {        int left = 0, n = str.size();        for (int i = 0; i <= n; ++i) {            if (i == n || str[i] == ' ') {                reverse(str, left, i - 1);                left = i + 1;            }        }        reverse(str, 0, n - 1);    }    void reverse(vector
      
       & str, int left, int right) {        while (left < right) {            char t = str[left];            str[left] = str[right];            str[right] = t;            ++left; --right;        }    }};
      
     

 

我们也可以使用C++ STL中自带的reverse函数来做，我们先把整个字符串翻转一下，然后再来扫描每个字符，用两个指针，一个指向开头，另一个开始遍历，遇到空格停止，这样两个指针之间就确定了一个单词的范围，直接调用reverse函数翻转，然后移动头指针到下一个位置，在用另一个指针继续扫描，重复上述步骤即可，参见代码如下：

 

解法二：

class Solution {public:    void reverseWords(vector
     
      & str) {        reverse(str.begin(), str.end());        for (int i = 0, j = 0; i < str.size(); i = j + 1) {            for (j = i; j < str.size(); ++j) {                if (str[j] == ' ') break;            }            reverse(str.begin() + i, str.begin() + j);        }    }};
     

 

类似题目：

 

参考资料：