Problem

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

第一种解法：将字符串转化为字符数组，用一头一尾两个指针向中间夹逼，遇到两个元音字母就进行位置交换。

第二种解法：相同的思路，一头一尾两个指针向中间夹逼。头指针是元音字母时，用尾指针找到最靠后的元音字母，并不断将靠后的元音字母加入StringBuilder；头指针是非元音字母时，就顺序放入非元音字母。这样，只用头指针放非元音字母，只用尾指针放元音字母。注意只有当头指针为元音字母时，才会操作尾指针。判断尾指针非元音字母的条件：while (j >= 0 && "AEIOUaeiou".indexOf(s.charAt(j)) == -1) j--;

Solution

1. Two Pointers --6ms

public class Solution {    static final String vowels = "aeiouAEIOU";    public String reverseVowels(String s) {        int first = 0, last = s.length() - 1;        char[] array = s.toCharArray();        while(first < last){            while(first < last && vowels.indexOf(array[first]) == -1){                first++;            }            while(first < last && vowels.indexOf(array[last]) == -1){                last--;            }            char temp = array[first];            array[first] = array[last];            array[last] = temp;            first++;            last--;        }        return new String(array);    }}

2. StringBuilder --15ms

public class Solution {    public String reverseVowels(String s) {        StringBuilder sb = new StringBuilder();        int j = s.length() - 1;        for (int i = 0; i < s.length(); i++) {            if ("AEIOUaeiou".indexOf(s.charAt(i)) != -1) {                while (j >= 0 && "AEIOUaeiou".indexOf(s.charAt(j)) == -1) j--;                sb.append(s.charAt(j));                j--;            }            else sb.append(s.charAt(i));        }        return sb.toString();    }}