【LeetCode】290. Word Pattern (Python實現)

參考鏈接：https://blog.csdn.net/coder_orz/article/details/51693647

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

方法一：就是建立兩個字典，進行雙向映射檢查。

class Solution(object):
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        
        str_list = str.strip().split()
        if len(str_list) != len(pattern):
            return "false"

        dictmap = {}
        mapval = {}
        for i in range(len(pattern.strip())):
            if pattern[i] in dictmap.keys():
                if str_list[i] != dictmap[pattern[i]]:
                    return "false"
            else:
                #這個地方是爲了雙向映射的原因
                if str_list[i] in mapval:
                    return "false"
                dictmap[pattern[i]] = str_list[i]
                mapval[str_list[i]] = pattern[i]
        return "true"

方法二：記錄每個字符和單詞出現的第一個位置，使得每個字符和單詞兩邊出現的第一次位置都相同時，就能保證雙射。

class Solution(object):
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        

        words = str.split(' ')
        if len(words) != len(pattern):
            return False
        if map(pattern.find, pattern) == map(words.index, words):
            return "true"
        else:
            return "false"

方法三：（'a'，‘dog’）類似這種建立映射體系，這個長度和pattern及str中去重後的長度是相等的。

class Solution(object):
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        
        words = str.split(' ')
        if len(words) != len(pattern):
            return False
        return len(set(pattern)) == len(set(words)) == len(set(zip(pattern, words)))