LeetCode290. Word Pattern

最新推荐文章于 2024-01-10 12:35:36 发布

原创最新推荐文章于 2024-01-10 12:35:36 发布 · 173 阅读

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本文深入探讨了模式匹配算法，通过具体实例展示了如何判断一个字符串是否遵循给定的模式。文章详细解释了使用split(), zip()和set()等方法进行字符串处理的技巧，以及如何利用字典来存储键值对，确保模式和字符串之间的全匹配。

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Example 1:

Input: pattern = "abba", str = "dog cat cat dog"

Output: true

Example 2:

Input:pattern = "abba", str = "dog cat cat fish"

Output: false

Example 3:

Input: pattern = "aaaa", str = "dog cat cat dog"

Output: false

Example 4:

Input: pattern = "abba", str = "dog dog dog dog"

Output: false

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

思路:和之前的一道题比较像,可以先用split()进行字符串分割,再用zip()方法把两个字符串放到一起 ,然后用set()进行比较.

需要注意的是需要添加and后的条件,否则example4会返回True

class Solution:
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        l = str.split(' ')
        return  len(set(zip(pattern, l))) == len(set(pattern)) == len(set(l)) and len(pattern) == len(l)

还有就是用字典的方法存储键值对.最后要判断value是否有重复,

class Solution:
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        dic={}
        strs = str.split(' ')
        if len(pattern) != len(strs):
            return False
        for char,word in zip(pattern,strs):
            if char in dic:
                if dic[char] != word:
                    return False
            else:
                dic[char]=word
        if len(dic.values()) != len(set(dic.values())):
            return False
        return True