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<char, unsigned int> crmap;
unordered_map<char, unsigned int> 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<char, vector<size_t>>
key 表示字符,value 是 key 出现的位置的集合