一起刷题吧

一、题意分析

输入：两个字符串
输出：如果第二个字符串与第一个字符串模式相同，则返回 true，否则返回 false
难度：简单
标签：哈希

示例1:
输入: pattern = “abba”, str = "dog cat cat dog"
输出: true

示例 2:
输入:pattern = “abba”, str = "dog cat cat fish"
输出: false

示例 3:
输入: pattern = “aaaa”, str = "dog cat cat dog"
输出: false

示例 4:
输入: pattern = “abba”, str = "dog dog dog dog"
输出: false

二、参考代码

这个题目就很简单啦，直接使用哈希表存储两者的对应关系就可以了，需要注意的是，需要使用两个哈希来存储对应关系，用一个哈希表会有问题，比如反例：

print(s.wordPattern("abba", "dog cat cat fish"))
print(s.wordPattern("abba", "dog dog dog dog"))

我的实现代码：

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        if not pattern:
            return not s
        if not s:
            return not pattern
        i = 0
        s = s.split(" ")
        if len(s) != len(pattern):
            return False
        mapping = {}
        rev = set()
        while i < len(pattern):
            if pattern[i] not in mapping:
                if s[i] not in rev:
                    mapping[pattern[i]] = s[i]
                    rev.add(s[i])
                else:
                    return False
            else:
                if mapping[pattern[i]] != s[i]:
                    return False
            i += 1
        return True

参考了下官方给的 python 代码示例，代码写得更加优雅，代码如下：

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        word2ch = dict()
        ch2word = dict()
        words = s.split()
        if len(pattern) != len(words):
            return False
        
        for ch, word in zip(pattern, words):
            # 更简洁
            if (word in word2ch and word2ch[word] != ch) or (ch in ch2word and ch2word[ch] != word):
                return False
            word2ch[word] = ch
            ch2word[ch] = word
    
        return True

# 作者：LeetCode-Solution
# 链接：https://leetcode-cn.com/problems/word-pattern/solution/dan-ci-gui-lu-by-leetcode-solution-6vqv/
# 来源：力扣（LeetCode）
# 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。