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?

Similar

136. Single Number

137. Single Number II

 

Explain:

In one's-complement (反码) representation, positive numbers are simply represented as themselves, and negative numbers are represented by the one's complement of their absolute value

In two's-complement (补码) representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value

[+1] = [00000001]_原 = [00000001]_反 = [00000001]_补

[-1]  = [10000001]_原 = [11111110]_反 = [11111111]_补

计算机使用补码进行计算

 

public class Solution {
    public int[] singleNumber(int[] nums) {
        // Pass 1 : 
        // Get the XOR of the two numbers we need to find
        int diff = 0;
        for (int num : nums) {
            diff ^= num;
        }
        // Get its last set bit which num1 is diff from num2
        diff &= -diff;

        // Pass 2 :
        int[] rets = {0, 0}; // this array stores the two numbers we will return
        for (int num : nums)
        {
            if ((num & diff) == 0) // the bit is not set // use ()
            {
                rets[0] ^= num;
            }
            else // the bit is set
            {
                rets[1] ^= num;
            }
        }
        return rets;
    }
}