349. Intersection of Two Arrays-优快云博客

本文链接：https://blog.youkuaiyun.com/teminusign/article/details/51941161

本文介绍了一种计算两个数组交集的方法，通过多种Python实现方案对比，包括使用集合操作简化代码，展示了不同方法的运行效率。

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349. Intersection of Two Arrays

Leetcode link for this question

Discription:

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:
- Each element in the result must be unique.
- The result can be in any order.

Analyze:

Code 1:

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        reli=[]
        for i in nums1:
            if i in nums2:
                reli.append(i)
        for i in nums2:
            if i in nums1:
                reli.append(i)
        reli=set(reli)
        return list(reli)

Submission Result:

Status: Accepted
Runtime: 108 ms
Ranking: beats 7.75%

Code 2:

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        reli=set(nums1) & set(nums2)
        print reli
        return list(reli)

Submission Result:

Status: Accepted
Runtime: 68 ms
Ranking: beats 38.72%

Code 3:

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        c1=collections.Counter(nums1)
        c2=collections.Counter(nums2)
        print c1.keys(),c2.keys()
        return [i for i in c1.keys() if i in c2.keys()]

Submission Result:

Status: Accepted
Runtime: 152 ms
Ranking: beats 3.11%

Code 4:

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        li = []
        for k in nums1:
            if k in nums2 and k not in li:
                li.append(k)    
        return li