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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);
+}