260. Single Number III

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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.

  1. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
1.先异或,得到的值是两个只出现一次的整数的异或值。
2.找出该异或值中第一个不为0的位置,即两个整数不相等的位。
3.根据该位将整数划分为两部分,两个整数分别位于两个部分。异或即可分别得到两个整数。

class Solution {public:vector<int> singleNumber(vector<int>& nums) {int xres = nums[0];for (int i = 1; i < nums.size(); i++){xres ^= nums[i];}int iOne = 0;for (int i = 0; i < 32; i++){if ((xres >> i) & 1){iOne = i;break;}}int res1 = 0, res2 = 0;for (int i = 0; i < nums.size(); i++){if ((nums[i] >> iOne) & 1){res1 ^= nums[i];}else{res2 ^= nums[i];}}vector<int> res;res.push_back(res1);res.push_back(res2);return res;}};


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