依旧是利用a^b^b==a的bit manipulation的特性。另一个方法则利用了将字母作为数组下标的方法。（从本质上来讲，在计算机里面，可以认为char就是int了，就是说可以把char当作int来用，如，作为数组下标。）
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.方法一：用HashMap
public class Solution { public char findTheDifference(String s, String t) { //char res=''; Map<Character,Integer> map = new HashMap<>(); for(char c : s.toCharArray()) map.put(c,map.getOrDefault(c,0)+1); for(char c : t.toCharArray()) { map.put(c,map.getOrDefault(c,0)-1); if(map.get(c)==-1) return c; } return 0; } }方法二：用数组。本质上和方法一相同。
for (int i = 0; i < 26; i++) alpha[i] = 0; for (char c : s.toCharArray()) alpha[ c - 'a' ]++; for (char c : t.toCharArray()) { //could do decrement first, then check but yeah if (--alpha[c - 'a'] < 0) return c; } return 0; 一定要学会 arr[ c - 'a' ]++ 这种用法。
方法三：异或。a^b^b==a
public char findTheDifference(String s, String t) { char c = 0; for (int i = 0; i < s.length(); ++i) { c ^= s.charAt(i); } for (int i = 0; i < t.length(); ++i) { c ^= t.charAt(i); } return c; }
