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Add C version for LeetCode349 using array
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@ -281,6 +281,38 @@ impl Solution {
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}
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}
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```
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C:
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```C
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int* intersection1(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize){
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int nums1Cnt[1000] = {0};
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int lessSize = nums1Size < nums2Size ? nums1Size : nums2Size;
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int * result = (int *) calloc(lessSize, sizeof(int));
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int resultIndex = 0;
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int* tempNums;
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int i;
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//Calculate the number's counts for nums1 array
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for(i = 0; i < nums1Size; i ++) {
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nums1Cnt[nums1[i]]++;
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}
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//Check if the value existing in nums1 count array
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for(i = 0; i < nums2Size; i ++) {
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if(nums1Cnt[nums2[i]] > 0) {
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result[resultIndex] = nums2[i];
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resultIndex ++;
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//Clear this count to avoid duplicated value
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nums1Cnt[nums2[i]] = 0;
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}
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}
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* returnSize = resultIndex;
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return result;
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}
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```
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## 相关题目
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* 350.两个数组的交集 II
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