349. Intersection of Two Arrays

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

Submission Result:

Status: Accepted
Runtime: 80 ms
Ranking: beats 21.92%

### 哈夫曼编码实现 哈夫曼编码是一种基于字符频率的压缩技术,通过构建一棵二叉树来生成最优前缀码。以下是针对字符串 `'The following code computes the intersection of two arrays'` 的 Python 实现过程。 #### 字符频率统计 首先需要统计输入字符串中每个字符的出现次数。这可以通过遍历字符串并记录每种字符的数量完成[^2]。 ```python from collections import Counter def calculate_frequencies(text): frequencies = Counter(text) return dict(frequencies) text = 'The following code computes the intersection of two arrays' frequencies = calculate_frequencies(text) print("Character Frequencies:", frequencies) ``` #### 构建哈夫曼树 利用字符及其对应的频率数据,可以按照以下方式构建哈夫曼树: 1. 创建一个优先队列(最小堆),其中每个节点表示一种字符以及其频率。 2. 不断取出两个具有最低频率的节点,创建一个新的内部节点作为它们的父亲,并将其频率设置为两子节点频率之和。 3. 将新节点重新加入优先队列,直到只剩下一个根节点为止。 ```python import heapq class HuffmanNode: def __init__(self, char, freq): self.char = char self.freq = freq self.left = None self.right = None def __lt__(self, other): return self.freq < other.freq def build_huffman_tree(freq_dict): priority_queue = [] for char, freq in freq_dict.items(): node = HuffmanNode(char, freq) heapq.heappush(priority_queue, node) while len(priority_queue) > 1: l
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