查找表_leetcode290

# 解题思路：字典解决其对应关系 20190302 找工作期间
#使用字典，pattern当key,str当value，形成配对


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

        s = str.split()

        if len(pattern) != len(s):
            return False

        return len(set(zip(pattern, s))) == len(set(pattern)) == len(set(s))


class Solution(object):
    def wordPattern(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        #使用字典，pattern当key,str当value，形成配对
        dic = {}
        strToList= str.split()
        if len(pattern) != len(strToList) or len(set(pattern)) != len(set(strToList)):
            return False
        for i, val in enumerate(pattern):
            if val not in dic:
                dic[val] = strToList[i]
            elif dic[val] != strToList[i]:
                return False
        return True

查找表_leetcode290

原文：https://www.cnblogs.com/lux-ace/p/10546937.html