Leetcode 349 两个数组的交集

　　简单问题，熟悉下 C 的哈希表，使用第三方库：

　　C:

typedef struct Hash
{
    int key;
    UT_hash_handle hh;
} Hash;

int hashExit(int key, Hash **hashs)
{
    Hash *target = NULL;
    HASH_FIND_INT(*hashs, &key, target);
    if (target == NULL)
        return 0;
    return 1;
}

int hashAdd(int key, Hash **hashs)
{
    Hash *target = (Hash *)malloc(sizeof(Hash));
    target->key = key;
    if (hashExit(key, hashs) == 1)
        return 0;
    HASH_ADD_INT(*hashs, key, target);
    return 1;
}

void hashClear(Hash **set)
{
    struct Hash *current, *tmp;
    HASH_ITER(hh, *set, current, tmp)
    {
        HASH_DEL(*set, current);
        free(current);
    }
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int *intersection(int *nums1, int nums1Size, int *nums2, int nums2Size, int *returnSize)
{
    Hash *nums1Hashs = NULL, *reHashs = NULL;
    int smallSize = nums1Size > nums2Size ? nums2Size : nums1Size;
    int *reArr = (int *)malloc(sizeof(int) * smallSize);
    int rePoint = 0;
    for (int i = 0; i < nums1Size; i++)
        hashAdd(nums1[i], &nums1Hashs);
    for (int i = 0; i < nums2Size; i++)
    {
        if (hashExit(nums2[i], &nums1Hashs) == 1)
        {
            if (hashAdd(nums2[i], &reHashs) == 1)
            {
                reArr[rePoint] = nums2[i];
                rePoint++;
            }
        }
    }
    hashClear(&nums1Hashs);
    hashClear(&reHashs);
    *returnSize = rePoint;
    return reArr;
}

　　JAVA:

class Solution {
         public final int[] intersection(int[] nums1, int[] nums2) {
            Set<Integer> nums1Set = new HashSet<Integer>();
            for (int i = 0; i < nums1.length; i++) nums1Set.add(nums1[i]);
            Set<Integer> reSet = new HashSet<Integer>();
            for (int i = 0; i < nums2.length; i++) {
                if (nums1Set.contains(nums2[i])) reSet.add(nums2[i]);
            }
            int[] reArr = new int[reSet.size()];
            int i = 0;
            Iterator<Integer> iterator = reSet.iterator();
            while (iterator.hasNext()) {
                reArr[i] = iterator.next();
                i++;
            }
            return reArr;
        }
}

　　JS:

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function (nums1, nums2) {
    let nums1Set = new Set(nums1), reSet = new Set(), reArr = [];
    for (let i = 0; i < nums1.length; i++) nums1Set.add(nums1[i]);
    let re = [], point = 0;
    for (let i = 0; i < nums2.length; i++) {
        if (nums1Set.has(nums2[i])) {
            if (reSet.has(nums2[i])) continue;
            reSet.add(nums2[i]);
            reArr.push(nums2[i]);
        }
    }
    return reArr;
};