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?
思路：将数组元素逐个异或。
举个栗子：2 3 4 2 3
所有数字依次异或运算：2 xor 3 xor 4 xor 2 xor 3 = (2 xor 2) xor (3 xor 3) xor 4= 0 xor 0 xor 4 = 4
最终结果4就是我们要找的那个只出现一次的数字。
转载于:https://www.cnblogs.com/vincent93/p/6686476.html
