LeetCode: 260. Single Number III

题目描述

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:

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

解题思路

先对每一位进行异或运算找到待求的两个数字不同的位，然后筛选出该位为不同值时对应出现一次的数字。

AC 代码

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        int mask = 1;
        int nor = 0;
        
        // 找到两个数字不同的位
        do
        {
            for(int num : nums)
            {
                nor ^= (num&mask);
            }
        }while(nor == 0 && (mask <<= 1));

        int ans0=0, ans1=0;
        // 该位为零的数，只有待求值中的一个是出现一次的
        for(int num : nums)
        {
            if((num&mask) == 0)
            {
                ans0 ^= num;
            }
        } 
        // 该位不为0的数，只有待求值中的一个是出现一次的
        for(int num : nums)
        {
            if((num&mask) != 0)
            {
                ans1 ^= num;
            }
        } 
        
        return {ans0, ans1};
    }
};