260. Single Number III

leetcode 260

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?

解题思路：

  数组nums中所有的数异或的结果res为只出现一次的两个数的异或值，res二进制的各个位中第一个为1的位置（由于1^0=1），这两个数一定是一个是1一个是0。所以根据这一位置是1还是0可以将数组nums划分成两组，分别在每一组中对所有数进行异或，即可得到要求的两个数。代码如下：

class Solution {public:    vector<int> singleNumber(vector<int>& nums) {        int res[2];res[0]=0;//保存数组中所有数的异或结果for(int i=0;i<nums.size();i++)res[0]=res[0]^nums[i];int mark=1;//非0位是异或结果中第一个1出现的位置for(i=1;i<=32;i++){if((res[0]&mark)==mark)break;mark=mark<<1;//当前第i位上非1，则进入下一次循环看第i+1位是否为1}//清零分别保存两组的异或结果res[0]=0;res[1]=0;for(i=0;i<nums.size();i++){if((nums[i]&mark)==mark)res[0]=res[0]^nums[i];elseres[1]=res[1]^nums[i];}vector<int> v(res,res+2);return v;    }};