交集运算如果数组 1 = { 1,2,3,4,6}
  数组 2 = {1,2,5,6,7 }
那么，数组1和数组2的交集是
array1 ^ array 2 = {1,2,3,4,6} ^ {1,2,5,6,7} = {1,2,6}
一组共同的元素被称为交集。
交集的逻辑如下 −
k=0;for(i=0;i<size1;i++){ for(j=0;j<size2;j++){ if(a[i]==b[j]){ intersection[k]=a[i]; k++; } }}
程序以下是执行两个数组交集操作的c程序 −
演示
#include<stdio.h>int removerepeated(int size,int a[]);void sort(int size,int a[]);main(){ int i,size1,size2,size,j=0,k,intersectionsize; printf("enter size of an array1
"); scanf("%d",&size1); printf("enter size of an array2
"); scanf("%d",&size2); int a[size1],b[size2],uni[size1+size2]; if(size1<size2){ intersectionsize=size1; }else if(size1>size2){ intersectionsize=size2; }else{ intersectionsize=size1; } int intersection[intersectionsize]; printf("enter numbers for array 1
"); for(i=0;i<size1;i++){ scanf("%d",&a[i]); } printf("enter numbers for array 2
"); for(i=0;i<size2;i++){ scanf("%d",&b[i]); } //intersection starts k=0; for(i=0;i<size1;i++){ for(j=0;j<size2;j++){ if(a[i]==b[j]){ intersection[k]=a[i]; k++; } } } //sorting sort(k,intersection); //removing size=removerepeated(k,intersection); printf("array after intersection
"); if(size>0){ for(i=0;i<size;i++){ printf("%d
",intersection[i]); } }else{ printf("no intersection
"); }}int removerepeated(int size,int a[]){ int i,j,k; for(i=0;i<size;i++){ for(j=i+1;j<size;){ if(a[i]==a[j]){ for(k=j;k<size;k++){ a[k]=a[k+1]; } size--; }else{ j++; } } } return(size);}void sort(int size,int a[]){ int i,j,temp; for(i=0;i<size;i++){ for(j=i+1;j<size;j++){ if(a[i]>a[j]){ temp=a[i]; a[i]=a[j]; a[j]=temp; } } }}
输出当上述程序被执行时，它产生以下结果 −
enter size of an array15enter size of an array22enter numbers for array 145678enter numbers for array 241array after intersection4
以上就是c程序实现对两个数组进行交集操作的详细内容。