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 Reverse Words in a String 没有区别，由于之前那道题就是用 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，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);

}

}

};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/186

 

类似题目：

Reverse Words in a String III

Reverse Words in a String

Rotate Array

 

参考资料：

https://leetcode.com/problems/reverse-words-in-a-string-ii/

https://leetcode.com/problems/reverse-words-in-a-string-ii/discuss/53851/Six-lines-solution-in-C%2B%2B

https://leetcode.com/problems/reverse-words-in-a-string-ii/discuss/53775/My-Java-solution-with-explanation

 

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