LeetCode136. Single Number

题目

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?

解答

思路：异或定义：相异为真，相同为假。相同数字异或为0，最后剩下的就是只出现过一次的。

class Solution {
    public int singleNumber(int[] nums) {
        int i = 1, result = nums[0];
        while (i < nums.length) result ^= nums[i++];
        return result;
    }
}