260. Single Number III

260. Single Number III

Medium

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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.

Example:

Input:  [1,2,1,3,2,5]
Output: [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: I have get tips from someone's idea, first XOR all the nums in array, you will get two numbers' XOR result, XOR number means the different bits of the two numbers, so we can just pick one to split the array into two, each will contains a final result.

 

class Solution {
    public int[] singleNumber(int[] nums) {
        int resultXOR = 0;
        for (int i = 0; i < nums.length; i++) {
            resultXOR ^= nums[i];
        }
        
        int oneBitXOR = 0;
        for (int i = 0; i < 32; i++) {
            if ((resultXOR & (1 << i)) != 0) {
                oneBitXOR = 1 << i;
                break;
            } 
        }
        int[] result = {0, 0};
        for (int i = 0; i < nums.length; i++) {
            if ((oneBitXOR & nums[i]) == 0) {
                result[0] ^= nums[i];
            } else {
                result[1] ^= nums[i];
            }
        }
        
        return result;
    }
}