检查数组是否是另一个数组的子集

Question

我有两个数组

1. 1, 3, 5, 7, 9
2. {3,5}或{1.9}。(从左到右顺序)

所以第二个数组是第一个数组的子集但如果第二个数组是{5.3}或{9.1}，则不是子集(从右到左顺序)

我的代码是

#include <stdio.h>
void subset(int set11[], int set12[])
{
int length1;
int set1[] = {1, 5, 3, 9, 7, 0, 5};
length1 = sizeof(set1) / sizeof(int);
int length2;
int set2[] = {5, 9};
length2 = sizeof(set2) / sizeof(int);
int i = 0;
int j = 0;
int count = 0;
for (i = 0; i < length1; i++)
{
if (set1[i] == set2[j])
{
count = 1;
}
}
printf(" is n");
if (count == 1)
{
printf("is subset");
}
else
{
printf("not subset");
}
}
int main()
{
int set11[] = {1, 5, 3, 9, 7};
int set12[] = {5, 9};
subset(set11, set12);
printf("");
return 0;
}

我得到所有情况下的输出，而不是子集

Answer 1

在逻辑上应用了一些变化。请参考评论。

#include <stdio.h>
#include <stdbool.h>
void subset(int set11[], int set12[])
{
int length1;
int set1[] = {1,3,5,7,9};
length1 = sizeof(set1) / sizeof(int);
int length2;
int set2[] = {3,5};
length2 = sizeof(set2) / sizeof(int);
int i = 0;
bool isSubset = true;   //will indicate weather the second array is subset or not
//    int j = 0;    //
int startPosition = 0;  // will decide the starting position for searching in  the main array.  {set 2}
//    int count = 0; //not needed; we will represent it with bool variable 'isSubset'.
for (i = 0; i < length2; i++)   //Iterating through the subset array
{
bool isFound = false;
for (int j=startPosition;j<length1;j++){        //Iterating through the original array {set 1}
if (set2[i]==set1[j]){  //if element from second array is found in first array then...
isFound = true;     //found the element
startPosition = j+1;        //increasing the starting position for next search in the first array.
printf("t%d found at %dn",set2[i],j);
break;
}
}
if(isFound==false){     //if not found then this is not subarray.
printf("t%d not foundn",set2[i]);
isSubset = false;
break;
}
}
//    printf(" is n");
if (isSubset)
{
printf("subset");
}
else
{
printf("not subset");
}
}
int main()
{
int set11[] = {1, 5, 3, 9, 7};
int set12[] = {5, 9};
subset(set11, set12);
printf("");
return 0;
}

Answer 2

您可以运行嵌套循环来获取子集数组中所有匹配的元素

for (i = 0; i < length1; i++) 
{
for(k = 0; k < length2; k++) 
{
if (set1[i] == set2[k])
{
count == 1;
}
}
}

外层循环用于for First数组检查子集/第二个数组

中任意位置的元素的内循环set1[] = {1,5,3,9,7,0,5};Set2 [] = {5,9};

如果我们运行一个嵌套循环，我们将得到所有的子集，而不管它们在第二个数组(set2)中的位置

无论set2是{5,9}还是{9,5}，计数器变量都会增加。