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.

解题思路 1

利用两个 hashtable，就能完成统计字符出现次数和出现的位置。

class Solution {public:    int firstUniqChar(string s) {        unordered_map
    
      crmap;        unordered_map
     
       cimap;        for (size_t i = 0; i < s.size(); i++) {            crmap[s[i]]++;            cimap[s[i]] = i;        }        char c = NULL;        string::iterator it = s.begin();        while (it != s.end()) {            if (crmap[*it] == 1) {                c = *it;                break;            }            it++;        }        return (c == NULL) ? -1 : cimap[c];    }};
     
    

解题思路 2

利用 hashtable 的变形

unordered_map
    
     >
    

key 表示字符，value 是 key 出现的位置的集合