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.(小写字母）

 

首先这里假设只有小写字母，自然想到开一个大小为26的数组，保存每个字母出现的次数，然后在扫描一次字符串，判断出现次数是否为1，若是则直接返回小标，扫描结束后若还没有返回，说明没有字符只出现一次，返回1.

 1 int firstUniqChar(char* s) {
 2     int m[26] = {0}, len = strlen(s), i;
 3     for(i = 0; i < len; i++){
 4         m[s[i] - 'a']++;
 5     }
 6     for(i = 0; i < len; i++){
 7         if(m[s[i] - 'a'] == 1)
 8             return i;
 9     }
10     return -1;
11 }

如果没有假设只包含小写字母，则可以用map

 1 class Solution {
 2 public:
 3     int firstUniqChar(string s) {
 4         int len = s.length(), i;
 5         map<char, int> map1;
 6         for(i = 0; i < len; i++){
 7             map1[s[i]]++;
 8         }
 9         for(i = 0; i < len; i++){
10            if(map1[s[i]] == 1)
11                 return i;
12         }
13         return -1;
14     }
15 };

 Leetcode 387 First Unique Character in a String_数组

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