leetcode 136. Single Number

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Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

我想了半天也没想到如何不使用格外空间又线性的解法，于是去看solutions。原来！原来用异或！！

public class Single_Number_136 {public int singleNumber(int[] nums) {int num=0;for(int i=0;i<nums.length;i++){num=num^nums[i];}return num;}public static void main(String[] args) {// TODO Auto-generated method stubSingle_Number_136 s=new Single_Number_136();int[] a=new int[]{1,2,3,1,3};System.out.println(s.singleNumber(a));}}

因为A XOR A = 0 ，并且 XOR operator 满足交换律，所以顺序不会影响结果。
异或的性质：
1、交换律
2、结合律（即(a^b)^c == a^(b^c)）
3、对于任何数x，都有x^x=0，x^0=x
4、自反性 A XOR B XOR B = A xor 0 = A

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