leetcode-2085.统计出现过一次的公共字符串

题目链接：2085. 统计出现过一次的公共字符串 - 力扣（LeetCode）

解题思路：

1、暴力破解

首先想到的是暴力破解，用两个循环遍历列表，然后将单词出现的情况都记录在一个字典里面。最后遍历字典找到满足条件的值。

class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        map1 = {}
        for wd in words1:
            if wd in map1:
                map1[wd] += 1
            else:map1[wd] = 1
        map2 = {}
        for wd in words2:
            if wd in map2:
                map2[wd] += 1
            else:map2[wd] = 1
        # print(map1,map2)
        count = 0
        for key, value in map1.items():
            if key in map2 and map1[key] == 1 and map2[key] == 1:
                count += 1
        
        return count

2、哈希

• counter()模块
  • 主要功能是支持方便、快速的计数，将元素统计，然后计数并且返回一个字典，键为元素，值为元素个数。结果的次数是从高到低的

    from collections import Counter
    
    list1 = ["a", "a", "a", "b", "c", "c", "f", "g", "g", "g", "f"]
    dic = Counter(list1)
    print(dic)
    #结果:次数是从高到低的
    #Counter({'a': 3, 'g': 3, 'c': 2, 'f': 2, 'b': 1})
    
    print(dict(dic))
    #结果:按字母顺序排序的
    #{'a': 3, 'b': 1, 'c': 2, 'f': 2, 'g': 3}
    
    print(dic.items()) #dic.items()获取字典的key和value
    #结果:按字母顺序排序的
    #dict_items([('a', 3), ('b', 1), ('c', 2), ('f', 2), ('g', 3)])
    
    print(dic.keys())
    #结果:
    #dict_keys(['a', 'b', 'c', 'f', 'g'])
    
    print(dic.values())
    #结果：
    #dict_values([3, 1, 2, 2, 3])
    
    print(sorted(dic.items(), key=lambda s: (-s[1])))
    #结果:按统计次数降序排序
    #[('a', 3), ('g', 3), ('c', 2), ('f', 2), ('b', 1)]
    
    for i, v in dic.items():
        if v == 1:
            print(i)
    #结果:
    #b
    

    解题代码

    class Solution:
        def countWords(self, words1: List[str], words2: List[str]) -> int:
            cnt1 = Counter(words1)
            cnt2 = Counter(words2)
            return sum(v == 1 and cnt2[w] == 1 for w,v in cnt1.items())