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.

思路：遍历每一个字母，在其位置后面的子序列进行查找，若有，则不唯一。同时把访问过的不唯一的字符进行存储，作为指示

代码1：

class Solution { public: int firstUniqChar(string s) { vector<char> yet; for(int i=0;i<s.size();i++){ if(find(yet.begin(),yet.end(),s[i])==yet.end()){ yet.push_back(s[i]); if( find(s.begin()+i+1,s.end(),s[i]) == s.end() ){ return (i); } } } return (-1); } };

代码2：

class Solution { public: int firstUniqChar(string s) { set<char> ss; for(int i=0;i<s.size();i++){ char ch=s[i]; if(ss.count(ch)){continue;} else{ss.insert(ch);} if((find(s.begin()+i+1,s.end(),ch)==s.end()) ){return i;} } return -1; } };

代码3：

class Solution { public: int firstUniqChar(string s) { vector<int> count(26, 0); for(int i=0;i<s.length();i++){ int cidx = s[i] - 'a'; //记录位置，为了防止，同时做标记，假如出现两次，即大于零（所以对于0位置，针对性加一），变为-1 if(count[cidx]==0) count[cidx] = i + 1; else count[cidx] = -1; } int idx = INT_MAX; for(int i=0;i<26;i++){ if(count[i]>0 && count[i]<idx){ idx = count[i]; } } return idx==INT_MAX?-1:idx-1; } };