问题:
Given a non-empty 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?
Example 1:
Input: [2,2,1] Output: 1
Example 2:
Input: [4,1,2,1,2] Output: 4
代码:
# -*- coding:utf-8 -*-
__author__ = 'yangxin_ryan'
"""
Soultions:
性质一:
交换律 a ^ b = b ^ a,
性质二:
a ^ a = 0。
于是我们可以将所有元素依次做异或操作,相同元素异或结果为0,
因此最终剩下的元素