[LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词之二
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：
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class Solution {public: void reverseWords(vector<char>& 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<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：
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class Solution {public: void reverseWords(vector<char>& 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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原文链接：https://www.cnblogs.com/grandyang/p/6703311.html
