260. Single Number III

题目

 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:

    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?

解析

// single nunmber iii
class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        
        vector<int > vec;

        int result = 0;
        for (int i = 0; i < nums.size();i++)
        {
            result ^= nums[i]; //改变了nums[0]的值，后续使用有问题
        }

        //两个不同的数在不同的位上，若数字不同，则相应的位为1
        if (result==0)
        {
            return vec;
        }

        //用flag找出num第一个不为0的位  
        int flag = 1;
        while ( (flag&result) == 0)  // (flag&result) == 0
        {
            flag = flag << 1;
        }

        int data1=0, data2=0; //0 异或任何数是保留原值的作用
        for (int j = 0; j < nums.size();j++)
        {
            if (nums[j]&flag) //某一位，相同的数与操作进同一分支
            {
                data1 ^= nums[j];
            }
            else
            {
                data2 ^= nums[j];
            }
        }
        vec.push_back(data1);
        vec.push_back(data2);

        return vec;
    }
};

题目来源

• 260. Single Number III