位操作-找到数组中只出现一次的数字

题目一：数组中数字都两两相同，只有一个不同，找出该数字：

int findUnique(int* a, int len){int i = 1;int temp = a[0];for(; i < len; i++){temp = temp ^ a[i];}printf("%d ", temp);} 

题目二：数组中数字两两相同，有两个不同，找出这两个：

#include <stdlib.h>int a[] = {1,1,2,4,3,3,2,5};int findXorSum(int* a, int len){int i = 0;int temp = 0;for(; i < len; i++){temp = temp ^ a[i];}return temp;} int findFirstBit1(int n)  {      int count = 1;      while(!( n & 1))      {          n = n>>1;          count++;      }      return count;  }  int isBit1(int a, int count){a = a >> count-1;return (a & 1);}void findTwoUnique(int* a, int len){int i = 0;int m = 0, n = 0;int temp = findXorSum(a, len);int count = findFirstBit1(temp);for(; i < len; i++){if(isBit1(a[i],count)){m = m ^ a[i];}else{n = n ^ a[i];}}printf("%d, %d", m, n);}int main(){findTwoUnique(a,8);}