[LeetCode][JavaScript]Single Number III

Single Number III

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

https://leetcode.com/problems/single-number-iii/

https://leetcode.com/discuss/52351/accepted-java-space-easy-solution-with-detail-explanations

第一轮遍历利用异或去掉出现两次的数，留下两个出现一次数的异或。

由于两个数不同，异或的结果中肯定有一位是1。

再遍历输入数组的时候，把输入的数与这一位做与运算，根据结果是否为零，可以把数组分为两份。

结果的两个数会被分开，异或去掉出现两次的数，最后得出答案。

 1 /**
 2  * @param {number[]} nums
 3  * @return {number[]}
 4  */
 5 var singleNumber = function(nums) {
 6     var two = 0, i;
 7     for(i = 0; i < nums.length; i++){
 8         two ^= nums[i];
 9     }
10     var pos = 1;
11     two = two.toString(2);
12     for(i = two.length - 1; i >= 0; i--){
13         if (two[i] === ‘1‘){
14             break;
15         }
16         pos *= 2;
17     }
18     var half1 = 0, half2 = 0;
19     for(i = 0; i < nums.length; i++){
20         if(nums[i] & pos){
21             half1 ^= nums[i];
22         }else{
23             half2 ^= nums[i];
24         }
25     }
26     return [half1, half2];
27 };