给两个数组：arr1[0..m-1] 和arr2[0..n-1]. 判断arr2是否是arr1的一个子集合，两个数组都是未排序的。
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、简单方法就是双重遍历
判断arr2矩阵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;}
但是对于重复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;}
