Leetcode 290. Word Pattern

Problem

Given a pattern and a string s, find if s 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 s.

Algorithm

Use the dictionary to save the two list.

Code

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        plen = len(pattern)
        dict_p = dict()
        dict_s = dict()
        word = ""
        index = 0
        for c in 服务器托管网s:
            if c >= 'a' and c  'z':
                word += c
            else:
                if index >= plen:
 服务器托管网                   return False
                pi = pattern[index]
                if index  plen and not pi in dict_p and not word in dict_s:
                    dict_p[pi] = word
                    dict_s[word] = pi
                elif pi in dict_p and dict_p[pi] != word or word in dict_s and dict_s[word] != pi:
                    return False
                index += 1
                word = ""
        
        if index >= plen:
            return False
        pi = pattern[index]
        if index  plen and not pi in dict_p and not word in dict_s:
            dict_p[pi] = word
            dict_s[word] = pi
        elif pi in dict_p and dict_p[pi] != word or word in dict_s and dict_s[word] != pi:
            return False

        for i in range(index, plen):
            if not pattern[i] in dict_p:
                return False
        return True

服务器托管，北京服务器托管，服务器租用 http://www.fwqtg.net

相关推荐: 前端学习笔记202310学习笔记第一百零七天-vue3-生命周期4

服务器托管，北京服务器托管网服务器托管网服务器托管，服务器租用 http://www.fwqtg.net 机房租用，北京机房租用，IDC机房托管， http://www.fwqtg.net相关推荐: 学正则表达式必用的工具正则可视化在线工具-更服务器托管网直观…