题目信息

• 链接: intersection-of-two-arrays

• Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must be unique and you may return the result in any order.

  Example 1:

  Input: nums1 = [1,2,2,1], nums2 = [2,2]
  Output: [2]
  
  

  Example 2:

  Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
  Output: [9,4]
  Explanation: [4,9] is also accepted.
  
  

  Constraints:

  • 1 <= nums1.length, nums2.length <= 1000
  • 0 <= nums1[i], nums2[i] <= 1000

• 描述:

  • 给定两个数组  nums1  和  nums2 ，返回_它们的交集_ 。输出结果中的每个元素一定是唯一的。我们可以不考虑输出结果的顺序。

解题思路

• 转成set取交集即可，set可以保证不重复和O(1)复杂度判断元素是否在set内
• 遍历全部num，找到即出现在nums1的元素又出现在nums2的num，放入result中
• 一行答案: list(set(nums1) & set(nums2))

代码

  class Solution:
	  def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
		  result = []
		  nums = nums1 + nums2
		  nums1 = set(nums1)
		  nums2 = set(nums2)
		  nums = set(nums)
		  for num in nums:
			  if num in nums1 and num in nums2:
				  result.append(num)
		  return result