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?
分析：
使用^即可，最后剩余的结果就是我们需要的。
/** * @param {number[]} nums * @return {number} */ var singleNumber = function(nums) { let result = 0; for(let i = 0; i <nums.length; i++) { result^= nums[i]; } return result; };
原文：http://www.cnblogs.com/gogolee/p/6657186.html