[leetcode]260. Single Number III

最新推荐文章于 2024-07-31 16:36:59 发布

xiaocong1990

最新推荐文章于 2024-07-31 16:36:59 发布

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本文介绍了一种线性时间复杂度的算法来找出数组中仅出现一次的两个元素，并提供了详细的实现步骤及代码示例。

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链接：https://leetcode.com/problems/single-number-iii/description/

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?

class Solution{
public:
    vector<int> singleNumber(vector<int>& nums)
    {
        vector<int> res(2,0);
        int mark=0;
        for(auto num:nums)
            mark^=num;
        
        //The statement "int lowbit=mark&(~(mark-1));" can find the rightmost set bit. For     
        //example, for mark = 4, bit = 2, i.e., lowbit is 
        //"00000000000000000000000000000010" in binary format  , the right first 1 in 
        //binary format
        int lowbit=mark&(~(mark-1));
        for(auto num:nums)
        {
            if((num&lowbit)==0)
                res[0]^=num;
            else
                res[1]^=num;

        }
        return res;
    }
};