标签：osi ediff code div let exp length abc ++
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
Example:
Input: s = "abcd" t = "abcde" Output: e Explanation: ‘e‘ is the letter that was added.
题目含义：给定两个字符串s和t，都只包含小写字母，字符串t是由字符串s打乱顺序且在额外的随机位置上添加一个字符组成，请找出这个字符
方法一：
1 public char findTheDifference(String s, String t) { 2 char result = 0; 3 for (int i=0;i<t.length();i++){ 4 result += t.charAt(i); 5 } 6 7 for (int i=0;i<s.length();i++){ 8 result -= s.charAt(i); 9 } 10 return result; 11 }
方法二：
1 public char findTheDifference(String s, String t) { 2 char result = 0; 3 for (int i=0;i<s.length();i++){ 4 result ^= s.charAt(i); 5 } 6 7 for (int i=0;i<t.length();i++){ 8 result ^= t.charAt(i); 9 } 10 return result; 11 }
标签：osi ediff code div let exp length abc ++
原文地址：http://www.cnblogs.com/wzj4858/p/7718870.html
