题目:
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.
翻译:
给定一个字符串,找到其中第一个没有重复的字符返回它的下标。如果不存在这样的字符,返回-1。
例子:
s = "leetcode" 返回 0. s = "loveleetcode", 返回 2.
思路:
先用map将字符串中出现的字符和它出现的次数进行映射,然后遍历一次map找到出现次数为1的那个字符,返回它的下标。
C++代码(Visual Studio 2017):
#include "stdafx.h" #include <iostream> #include <string> #include <map> using namespace std; class Solution { public: int firstUniqChar(string s) { map<char, int> m; for (int i = 0; i < s.size(); i++) { m[s[i]]++; } for (int i = 0; i < s.size(); i++) { if (m[s[i]] == 1) return i; } return -1; } }; int main() { Solution s; string str = "loveleetcode"; int result; result = s.firstUniqChar(str); cout << result; return 0; }