【前端算法】只出现一次的数字 II,位运算符：NOT，AND 和 XOR

示例 1:
输入: [2,2,3,2]
输出:3

示例 2:
输入:[0,1,0,1,0,1,99]
输出:99

/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function(nums) {
	var obj = {};
	//1.统计每个元素出现的次数
	for (var i = 0; i < nums.length; i++) {
		var key = nums[i];
		if (obj[key]) {
			obj[key]++;
		} else {
			obj[key] = 1;
		}
	}
	//2.找出只出现一次的元素
	for (var k in obj) {
		if (obj[k] === 1) {
			return k
		}
	}
};

5 & 1 = 1

5| 1 = 5

 ~5 = -6

/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function(nums) {
	var ones = 0,
		twos = 0;
	for (var i = 0; i < nums.length; i++) {
		ones = ones ^ nums[i] & ~twos;
		twos = twos ^ nums[i] & ~ones;
	}
	return ones
};