The "C language" output has 2 occurrences of 1 numbers in a set of numbers

Before we wrote a function, written in a group of numbers have a number only appeared once, the other books have appeared 2 times, the idea is very simple, directly in turn or can find this number. See my previous blog post for details.

Then let's take a look at the present question.

The number of outputs in a group of 2 appears 1 times. Then the remaining number appeared 2.

Our idea of this problem is as follows:

1. Differences or results of different numbers

2. Find the number of digits in the XOR result that 1 appears.

3. Group XOR, when the occurrence of 1 of the bit XOR, not 1 of the bit XOR, the original array can be divided into 2 groups, and then different or will be able to get 2 only one occurrence of the number.

It's easy to understand the code:

#include <stdio.h>void getnumber (int *arry,int *numone, int *numsecound,int  len) {             int i =  0;             int pivot =  0;             int pos =  1;             for (i = 0;  i < len; i++)             {                          pivot ^= * (arry+i);             }              while (!) ( pivot&1))             {                         pivot =  pivot>>1;                         pos++;             }              for (i = 0; i<len;i++)              {                          if ((* (arry+i) >>pos-1)  & 1)                           {                                     *numone ^ = * (arry+i);                         }                          else                          {                                      *numsecound ^= * (arry+i);                         }             }}int main () {                        int num1 =  0;             int num2 =  0;             int arry[] =  {33,19,20,8,37,8,37,5,25,33,19,20};             getnumber (arry,&num1,&num2, sizeof (Arry)/sizeof  (arry[0]);             printf (  "%d ", NUM1);             printf (  "%d", num2);              return 0;} 

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The "C language" output has 2 occurrences of 1 numbers in a set of numbers