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[题目]
Given a string, find the first non-repeating character in it and return it‘s index. If it doesn‘t exist, return -1.
Examples:
s = "leetcode" return 0. s = "loveleetcode", return 2.
Note: You may assume the string contain only lowercase letters.
[题目解析] 根据题意，思路比较简单，遍历字符串，用map存储对应字符以及出现次数，再次遍历原字符串，当出现第一个出现一次的字符时，返回对应的index即可，考虑到Note中提示的字符串只考虑小写字母组成的情况，可以用一个26位的数组表示map结构，代码如下。
public int firstUniqChar(String s) {
if(null == s || 0 == s.length() ) return -1;
int[] hash = new int[26];
char[] array = s.toCharArray();
for(int i = 0; i < array.length; i++){
int num = array[i] - ‘a‘;
hash[num]++;
}
for(int i = 0; i < array.length; i++){
int num = array[i] - ‘a‘;
if(hash[num] == 1){
return i;
}
}
return -1;
}
[LeetCode] NO. 387 First Unique Character in a String
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原文地址：http://www.cnblogs.com/zzchit/p/5831617.html
