From 3f7dd67a8031663d2b673a41587e3a640e0a4b1f Mon Sep 17 00:00:00 2001
From: FrankLin <linliawxm@gmail.com>
Date: Fri, 1 Apr 2022 10:57:25 -0400
Subject: [PATCH] Add C version for LeetCode349 using array

---
 problems/0349.两个数组的交集.md | 32 ++++++++++++++++++++++++++
 1 file changed, 32 insertions(+)

diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md
index 82be1829..64d80a37 100644
--- a/problems/0349.两个数组的交集.md
+++ b/problems/0349.两个数组的交集.md
@@ -281,6 +281,38 @@ impl Solution {
     }
 }
 ```
+
+C:
+```C
+int* intersection1(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize){
+
+    int nums1Cnt[1000] = {0};
+    int lessSize = nums1Size < nums2Size ? nums1Size : nums2Size;
+    int * result = (int *) calloc(lessSize, sizeof(int));
+    int resultIndex = 0;
+    int* tempNums;
+    
+    int i;
+
+    //Calculate the number's counts for nums1 array
+    for(i = 0; i < nums1Size; i ++) {
+        nums1Cnt[nums1[i]]++;
+    }
+
+    //Check if the value existing in nums1 count array
+    for(i = 0; i < nums2Size; i ++) {
+        if(nums1Cnt[nums2[i]] > 0) {
+            result[resultIndex] = nums2[i];
+            resultIndex ++;
+            //Clear this count to avoid duplicated value
+            nums1Cnt[nums2[i]] = 0;
+        }
+    }
+    * returnSize = resultIndex;
+    return result;
+}
+```
+
 ## 相关题目
 
 * 350.两个数组的交集 II