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<char>& str) {
int left = , n = str.size();
for (int i = ; i <= n; ++i) {
if (i == n || str[i] == ' ') {
reverse(str, left, i - );
left = i + ;
}
}
reverse(str, , n - );
}
void reverse(vector<char>& 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<char>& str) {
reverse(str.begin(), str.end());
for (int i = , j = ; i < str.size(); i = j + ) {
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：
参考资料：