350. Intersection of Two Arrays II

Problem

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.

https://leetcode.cn/problems/intersection-of-two-arrays-ii/

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.

Constraints:

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

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1’s size is small compared to nums2’s size? Which algorithm is better?
• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

Test Cases

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class Solution:
    def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:

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import pytest

from solution import Solution


@pytest.mark.parametrize('nums1, nums2, expected', [
    ([1,2,2,1], [2,2], [2,2]),
    ([4,9,5], [9,4,9,8,4], [9,4]),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums1, nums2, expected):
    result = sol.intersect(nums1, nums2)
    assert sorted(result) == sorted(expected)

Thoughts

349. Intersection of Two Arrays 的进阶版，求交集的时候要考虑词频。

把集合改为字典，值为词频。

Python 的 collections.Counter 可以直接求词频的交集。

Code

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from collections import defaultdict


class Solution:
    def intersect(self, nums1: list[int], nums2: list[int]) -> list[int]:
        min2 = lambda a, b: a if a <= b else b

        d1 = defaultdict(int)
        for num in nums1:
            d1[num] += 1

        d2 = defaultdict(int)
        for num in nums2:
            d2[num] += 1

        result = []
        for num, f1 in d1.items():
            if num in d2:
                f2 = d2[num]
                result.extend([num] * min2(f1, f2))

        return result

参考资料

• 本文引用

  • 349. Intersection of Two Arrays

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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