标签：int com bsp bool 一点 size == public code
原题地址：
https://leetcode.com/problems/reverse-vowels-of-a-string/description/
题目：
Write a function that takes a string as input and reverse only the vowels of a string.
Example 1:
Given s = "hello", return "holle".
Example 2:
Given s = "leetcode", return "leotcede".
Note:
The vowels does not include the letter "y".
解法：
我能想到的有两种解法：
（1）
先遍历一次字符串的每个字符，把元音字母的位置放在一个数组里面。然后按照数组将元音字母的位置交换即可。
class Solution { public: string reverseVowels(string s) { vector<int> m; for (int i = 0; i < s.size(); i++) { if (s[i] == ‘a‘ || s[i] == ‘e‘ || s[i] == ‘i‘ || s[i] == ‘o‘ || s[i] == ‘u‘ || s[i] == ‘A‘ || s[i] == ‘E‘ || s[i] == ‘I‘ || s[i] == ‘O‘ || s[i] == ‘U‘) { m.push_back(i); } } if (m.size() == 0) return s; for (int i = 0; i <= (m.size() - 1) / 2; i++) { char temp = s[m[i]]; s[m[i]] = s[m[m.size() - i - 1]]; s[m[m.size() - i - 1]] = temp; } return s; } };
（2）
利用两个变量i和j，一加一减，遇到元音的位置就停下来，然后交换。我认为这种方法比较巧妙一点。类似于这种i和j变量的应用的题目，还有Intersection of Two Arrays II这道题，可以联系起来总计一下这种变量的使用。
class Solution { public: bool isVowel(char c) { if (c == ‘a‘ || c == ‘A‘ || c == ‘e‘ || c == ‘E‘ || c == ‘i‘ || c == ‘I‘ || c == ‘o‘ || c == ‘O‘ || c == ‘u‘ || c == ‘U‘) return true; else return false; } string reverseVowels(string s) { int i = 0, j = s.size() - 1; while (i < j) { while (!isVowel(s[i]) && i < j) i++; while (!isVowel(s[j]) && i < j) j--; char temp = s[i]; s[i] = s[j]; s[j] = temp; i++; j--; } return s; } };
[LeetCode]345 Reverse Vowels of a String
标签：int com bsp bool 一点 size == public code
原文地址：http://www.cnblogs.com/fengziwei/p/7591353.html
