概述
[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:
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:
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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