260. Single number III

Given An array of numbers nums , in which exactly-elements appear only once and all the other elements appear exactly Twice. Find the 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. The above example, is [5, 3] also correct.

Your algorithm should run in linear runtime complexity. Could implement it using only constant space complexity?

1 /**2 * Return An array of size *returnsize.3 * Note:the returned array must is malloced, assume caller calls free ().4  */5 int* Singlenumber (int* Nums,intNumssize,int*returnsize) {6     intFlag =0;7     inti;8     intK =1;9     int*Res;Tenres = (int*)malloc(2*sizeof(int));//It must be malloc or not. Oneres[0] = res[1] =0; A      for(i =0; i < numssize; i++)//first different or out of these 2 numbers -Flag ^=Nums[i]; -      while(1) the     {  -         if(k & Flag)//find the first bit of 1 in binary, then these 2 digits, one of which is 1, the other is 0 -              Break;  -K = k <<1; +     } -      for(i =0; i < numssize; i++) +     { A         if(K & Nums[i])//put all of this in a 1-bit category, atres[0] ^= Nums[i];//different or out this is 0 of the number -         Else                                               //In the same place as the 1 number -res[1] ^=Nums[i]; -          -     } -*returnsize =2; in     returnRes; -}

260. Single number III