LeetCode 136 Single Number

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?

数组中的数字按顺序异或，最后剩下的就是独一无二的数字。

1234567

 public int singleNumber(int[] nums) {        int r = nums[0], l = nums.length;for (int i = 1; i < l; i++) {r = r ^ nums[i];}return r;    }