給兩個數組：arr1[0..m-1] 和arr2[0..n-1]. 判斷arr2是否是arr1的一個子集合，兩個數組都是未排序的。

例子：

Input: arr1[] = {11, 1, 13, 21, 3, 7}, arr2[] = {11, 3, 7, 1}
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {1, 2, 3, 4, 5, 6}, arr2[] = {1, 2, 4}
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {10, 5, 2, 23, 19}, arr2[] = {19, 5, 3}

Output: arr2[] is not a subset of arr1[]

1、簡單方法就是雙重遍歷

判斷arr2矩陣中的所有元素是不是在arr1

#include<stdio.h> bool isSubset(int arr1[], int arr2[], int m, int n) {int i = 0;int j = 0;for (i=0; i<n; i++){for (j = 0; j<m; j++){if(arr2[i] == arr1[j])break;}if (j == m)return 0;}return 1; }int main() {int arr1[] = {11, 1, 13, 21, 3, 7};int arr2[] = {11, 3, 7, 1};int m = sizeof(arr1)/sizeof(arr1[0]);int n = sizeof(arr2)/sizeof(arr2[0]);if(isSubset(arr1, arr2, m, n))printf("arr2[] is subset of arr1[] ");elseprintf("arr2[] is not a subset of arr1[]"); return 0; }
但是對于重復的元素方法好像不行， 例如{1, 4, 4, 2} 并不是 {1, 4, 2}的子集

bool isSubset(int arr1[], int arr2[], int m, int n) {int i = 0, j = 0;if(m < n)return 0;quickSort(arr1, 0, m-1);quickSort(arr2, 0, n-1);while( i < n && j < m ){if( arr1[j] <arr2[i] )j++;else if( arr1[j] == arr2[i] ){j++;i++;}else if( arr1[j] > arr2[i] )return 0;}if( i < n )return 0;elsereturn 1; }

總結

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