[LeetCode 260] 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?

Solution:

1. Use XOR to store the difference among these numbers, if XOR all elements, the result is the difference between two number like result = 3^5

then 3^result = 5 , 5^ result = 3

2. Find one digit 1 in the result, which can be used to distinguish 3 and 5. depends on this, XOR elements will be equal to 3^result = 5 , 5^ result = 3

O(n) use constant space

public int[] singleNumber(int[] nums) {
        int[] res = new int[2];
        int result = nums[0];
        for(int i=1;i