[算法相關] 260. Single Number III -取出非重複數字（加強版）

 

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?

 題意，如上所示，數組中非重複數字有兩個，現在需要找到這兩個元素。

首先給出思路：

 

If you were stuck by this problem, it's easy to find a solution in the discussion. However, usually, the solution lacks some explanations.

I'm sharing my understanding here:

The two numbers that appear only once must differ at some bit, this is how we can distinguish between them. Otherwise, they will be one of the duplicate numbers.

Let’s say the at the ith bit, the two desired numbers differ from each other. which means one number has bit i equaling: 0, the other number has bit i equaling 1.

Thus, all the numbers can be partitioned into two groups according to their bits at location i. the first group consists of all numbers whose bits at i is 0. the second group consists of all numbers whose bits at i is 1.

Notice that, if a duplicate number has bit i as 0, then, two copies of it will belong to the first group. Similarly, if a duplicate number has bit i as 1, then, two copies of it will belong to the second group.

by XoRing all numbers in the first group, we can get the first number. by XoRing all numbers in the second group, we can get the second number.

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
        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
            {
                rets[0] ^= num;
            }
            else // the bit is set
            {
                rets[1] ^= num;
            }
        }
        return rets;
    }
}

位運算簡直博大精深……

原文鏈接：https://leetcode.com/problems/single-number-iii/