c++之数组篇2：计算两个有序数组的交集

#include <iostream>
//using namespace std;

int mixed(int arr1[],int n1,int arr2[],int n2,int* mix )
{
int i = 0;
int j = 0;
int k = 0;
//mix = new int[n1];
while (i < n1 && j < n2)
{
if(arr1[i] == arr2[j])
{
mix[k++] = arr1[i];
//k++;
i++;
j++;
}
else if (arr1[i]>arr2[j])
j++;
else if (arr1[i]<arr2[j])
i++;
}
return k;
}
int main()
{
int *m = NULL; //存储相同元素的指针
int num;//存储两个数组相同元素个数
int a[] = {0,1,2,3,4};
int b[] = {1,3,5,7,9};
int len1 = sizeof(a)/sizeof(a[0]);
int len2 = sizeof(b)/sizeof(b[0]);
m = new int[len1];
num = mixed(a,len1,b,len2,m);
printf("两数组相同元素个数： %d\n",num);
printf("相同元素为：\n");
for(int i = 0;i < num;i++)
printf("%d",m[i]);

return 0;
}