All the numbers in an array are paired. Only one or two numbers are separate, and this number is calculated

If there is only one number, then we can use the bitwise OR (^) to get, the code is as follows:

#include <stdio.h>int main () {int arr[] = {1, 1, 2, 2, 3, 3, 4};int size = sizeof (arr)/sizeof (int); int num = 0;for (int i = 0; i < size; i++) {num ^= arr[i];} printf ("%d\n", num); system ("pause"); return 0;}

If it is two numbers, then we can split the array, and then respectively with 0 or, its split thought is the array of all the number of all or, the resulting result is the two individual occurrences of the same or the result, because or its nature is the same as 0, different 1, So the number of each of the 1 bits is its different bits, we can find it first for 1 of the bit to mark, the same or together for one, 0 or together, the result is two numbers, the implementation code is as follows:

#include <stdio.h>void find_num (int *arr, int size)//Its implementation function is encapsulated, because is to return two values, C language                                          //realize the two values are printed out {int num1 = 0;int num2 = 0;int end_num = 0; int flag = 0;for  (int i = 0; i < size; i++) {End_num  ^= * (arr + i);} while  (! ( end_num & 1)) {flag++;end_num >>= 1;} for  (int i = 0; i < size; i++) {Int tmp = arr[i]  >> flag;if  (tmp & 1) {num1 ^= arr[i];} Else{num2 ^= arr[i];}} printf ("num1=%d\nnum2=%d\n",  num1, num2);} Int main ()//validation phase {Int arr[] = { 1, 1,&nbsP;2, 2, 3, 3, 4, 5 };int size = sizeof (arr)  / sizeof ( Arr[0]); Find_num (arr,size); System ("pause"); return 0;}

If there is anything wrong, hope to criticize

This article is from the "Pawnsir It Road" blog, so be sure to keep this source http://10743407.blog.51cto.com/10733407/1713119

All the numbers in an array are paired. Only one or two numbers are separate, and this number is calculated