Description: Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
Link: 350. Intersection of Two Arrays II
Examples:
Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9] Explanation: [9,4] is also accepted.
思路: 发现一样的元素就放回返回值的list中并在原list中删除, pop(指定位置)， remove(指定元素)， del(a[0])，指定位置，但不传入index.
class Solution(object): def intersect(self, nums1, nums2): """ :type nums1: List[int] :type nums2: List[int] :rtype: List[int] """ res = [] if len(nums2) > len(nums1): nums1, nums2 = nums2, nums1 for i in nums1: if len(nums2) == 0: return res if i in nums2: res.append(i) nums2.remove(i) return res
日期: 2021-04-02 加油！
Leetcode 350. Intersection of Two Arrays II
原文：https://www.cnblogs.com/wangyuxia/p/14610344.html