哈希表-两个数组的交集

解法：由于同一个list中可能存在重复元素，所以考虑采用计数的方式解决问题，具体思路如下：

首先构建一个dict来记录list中的元素个数，称为count_dict；count_dict中的元素是一个长度为2的list，分别存储nums1和nums2中元素i的个数

class Solution:
    def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
        count_dict = {}     
        for i in nums1:
            if i not in count_dict:
                count_dict[i] = [1, 0]
            else:
                count_dict[i][0] += 1
        for i in nums2:
            if i not in count_dict:
                # nums2中有但nums1中没有的元素，就没必要再做记录了
                continue
            count_dict[i][1] += 1
            result_list = []
        for key in count_dict:
            # 遍历，统计某个元素i在nums1和nums2中出现次数的较小值num，若num不为0，则将num个元素i加入到结果list（result_list）中
            num = min(count_dict[key])
            if num:
                result_list += [key]*num
        return result_list

　

 列表还支持拼接操作

 

 

class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        count_dict = {}
        for i in nums1:
            if i not in count_dict:
                count_dict[i] = 1

        new_arr = []
        for i in nums2:
            if count_dict.get(i):
                new_arr.append(i)
                count_dict[i] -= 1
        return new_arr

count_dict[i] -= 1是为了count_dict.get(i)判断用
如果不减1，就会出现重复
如：

nums1 = [4,9,5, 4]
nums2 = [9,4,9,8,4]