from typing import List


class Solution:
	def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
		# 記錄兩個(gè)數(shù)組中每個(gè)數(shù)字出現(xiàn)的次數(shù)
		result1 = {}
		result2 = {}
		
		# 遍歷兩個(gè)數(shù)組，然后將其數(shù)字出現(xiàn)次數(shù)添加入字典中
		for i in nums1:
			if i not in result1.keys():
				result1[i] = 1

			else:
				result1[i] += 1

		for i in nums2:
			if i not in result2.keys():
				result2[i] = 1
			else:
				result2[i] += 1

		# 將兩個(gè)字典的鍵取出
		k1 = [i for i in result1.keys()]
		k2 = [i for i in result2.keys()]
		# 求兩個(gè)字典鍵的交集
		k = set(k1) & set(k2)

		# 結(jié)果列表
		r = []
		
		# 同一個(gè)鍵，代表兩個(gè)列表中都有出現(xiàn)，然后取最小的值為出現(xiàn)的次數(shù)，然后用列表的extend方法添加入結(jié)果列表中
		for i in k:
			m = min(result1[i], result2[i])
			r.extend([i]*m)

		return r

if __name__ == "__main__":
	s = Solution()
	nums1 = [1,2,2,1]
	nums2 = [2,2]
	r1 = s.intersect(nums1, nums2)
	print(r1)


            
          

給定兩個(gè)數(shù)組，編寫一個(gè)函數(shù)來(lái)計(jì)算它們的交集。

感覺(jué)只要用字典，速度都不會(huì)慢，因?yàn)樽值洳檎冶闅v都好快