python algorithm diary (string series) _leetcode 290. word law

Given a law and a pattern string str, str determines whether to follow the same rule.

Here follow the exact match means, for example, there is a corresponding pattern regularity in two-way connection between each letter string str and each non-empty word.

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

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/word-pattern

Establish a hash table mapping

class Solution:
    def wordPattern(self, pattern: str, str: str) -> bool:
        word_list = str.split()
        if len(set(pattern))!= len(set(word_list)): #避免"abba","dog dog dog dog"
            return False
        if len(pattern)!=len(word_list): #长度不等，模式不同
            return False
        str_map = {}
        for i,val in enumerate(pattern):
            if val not in str_map.keys():
                str_map[val]=word_list[i]  #建立字母与单词的映射
            if str_map[val]==word_list[i]:
                    continue
            else:
                    return False
        return True

LeetCode the comments of two lines of code from @ Miao Xiaowei  https://leetcode-cn.com/u/mou-xiao-wei/

The original map can also use this:

class Solution:
    def wordPattern(self, pattern: str, str: str) -> bool:
        res=str.split()
        return list(map(pattern.index, pattern))==list(map(res.index,res))
        # pattern.index('a')返回pattern字符串中字符‘a’首次出现的位置

Two two comparison pattern and the words are equal or unequal while at the same time:

class Solution:
    def wordPattern(self, pattern: str, str: str) -> bool:
        word_list = str.split()
        if len(set(pattern))!= len(set(word_list)):
            return False
        if len(pattern)!=len(word_list):
            return False
        for i in range(len(pattern)-1):
            if pattern[i]==pattern[i+1] and word_list[i]!=word_list[i+1]:
                return False
            elif pattern[i]!=pattern[i+1] and word_list[i]==word_list[i+1]:
                return False
        return True