Leetcode 387 First Unique Character in a String

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.

int firstUniqChar(char* s) {
    int m[26] = {0}, len = strlen(s), i;
    for(i = 0; i < len; i++){
        m[s[i] - 'a']++;
    }
    for(i = 0; i < len; i++){
        if(m[s[i] - 'a'] == 1)
            return i;
    }
    return -1;
}

如果没有假设只包含小写字母，则可以用map

class Solution {
public:
    int firstUniqChar(string s) {
        int len = s.length(), i;
        map<char, int> map1;
        for(i = 0; i < len; i++){
            map1[s[i]]++;
        }
        for(i = 0; i < len; i++){
           if(map1[s[i]] == 1)
                return i;
        }
        return -1;
    }
};