查找数问题

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1. 一个数组中，只有两个数出现一次，其他都出现两次，找出这两个出现一次的数.(异或的运用和位操作)

int get_xor_result(int a[],int length){    int tem = 0;    for(int i =0; i<length; i++)    {        tem ^= a[i];    }    return tem;}int get_pos(int tem){    for(int i = 0; i<32; i++)    {        if(tem & 1 == 1)            return i;        tem >>= 1;    }    return 0;}int get_num(int a[],int length,int *num1,int *num2){    int result,pos;    result = get_xor_result(a,length);    pos = get_pos(result);    *num1 = *num2 = 0;    for(int i=0; i<length; i++)    {        if((a[i]>>pos & 1) == 1)            *num1 ^= a[i];        else            *num2 ^= a[i];    }    return 0;}int main(void){    int a[10]={1,2,3,4,5,6,4,3,2,1};    int num1,num2;    get_num(a,10,&num1,&num2);    printf("%d,%d\n",num1,num2);    printf("hello world\n");    return 0;}

2.一个大数组中只有两个数有重复的，其他都是出现一次，找出是哪两个数？

#include <stdio.h>#define LENGTH 10static char array[LENGTH / 8 + 1];void get_num(int a[], int length, int *num1, int *num2){int tem = 0;*num1 = *num2 = 0;for (int i = 0; i < length; i++){tem = (array[a[i] / 8] >> (a[i] % 8)) & 1;if (tem != 1)array[a[i] / 8] |= 1 << a[i] % 8;else{if (*num1 == *num2)*num1 = a[i];else*num2 = a[i];}}}int main(){int num1, num2;int a[LENGTH] = { 1, 2, 11, 4, 11, 5, 6, 6, 7, 8 };get_num(a, LENGTH, &num1, &num2);printf("%d,%d\n", num1, num2);printf("hello world\n");return 0;}

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