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.

?思路：掃一遍，用array記錄char的重復次數。掃第二遍，碰到第一個char的個數是1就輸出，掃完沒有就是－1.

public class Solution {public int firstUniqChar(String s) {if(s.length()==0){return -1;}int[] res=new int[128];for(int i=0;i<s.length();i++){if(res[s.charAt(i)]==0){res[s.charAt(i)]=1;}else{res[s.charAt(i)]++;}}for(int i=0;i<s.length();i++){if(res[s.charAt(i)]==1){return i;}}return -1;} }

?

轉載于:https://www.cnblogs.com/Machelsky/p/5916541.html

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