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// 350. Intersection of Two Arrays II
#include "leetcode.h"
/*
* given two integer arrays 'nums1' and 'nums2' return an array of their
* intersection. each element in the result must appear as many times as it
* shows in both arrays and you may return the result in any order.
*/
int *intersect(int *nums1, int nums1Size, int *nums2, int nums2Size,
int *returnSize) {
int *hash = (int *)malloc(1001 * sizeof(int)),
*ans = (int *)malloc(1000 * sizeof(int)), n = 0;
for (int i = 0; i < nums1Size; i++)
hash[nums1[i]]++;
for (int i = 0; i < nums2Size; i++)
if (hash[nums2[i]]) {
ans[n++] = nums2[i];
hash[nums2[i]]--;
}
*returnSize = n;
free(hash);
return ans;
}
int main() {
int n11[] = {1, 2, 2, 1}, n21[] = {2, 2}, n12[] = {4, 9, 5},
n22[] = {9, 4, 9, 8, 4}, rs1, rs2;
int *i1 = intersect(n11, ARRAY_SIZE(n11), n21, ARRAY_SIZE(n21), &rs1);
int *i2 = intersect(n12, ARRAY_SIZE(n12), n22, ARRAY_SIZE(n22), &rs2);
for (int i = 0; i < rs1; i++)
printf("%d ", i1[i]); // expect: 2 2
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%d ", i2[i]); // expect: 4 9
printf("\n");
free(i1), free(i2);
}
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