290. 单词规律

English Version

题目描述

给定一种规律 pattern 和一个字符串 str ，判断 str 是否遵循相同的规律。

这里的 遵循 指完全匹配，例如， pattern 里的每个字母和字符串 str 中的每个非空单词之间存在着双向连接的对应规律。

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

说明:
你可以假设 pattern 只包含小写字母， str 包含了由单个空格分隔的小写字母。    

解法

Python3

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        ch2str, str2ch = {}, {}
        ss = s.split(' ')
        n = len(pattern)
        if n != len(ss):
            return False
        for i in range(n):
            if ch2str.get(pattern[i]) is not None and ch2str.get(pattern[i]) != ss[i]:
                return False
            if str2ch.get(ss[i]) is not None and str2ch.get(ss[i]) != pattern[i]:
                return False
            ch2str[pattern[i]] = ss[i]
            str2ch[ss[i]] = pattern[i]
        return True

Java

class Solution {
    public boolean wordPattern(String pattern, String s) {
        Map<Character, String> ch2str = new HashMap<>();
        Map<String, Character> str2ch = new HashMap<>();
        String[] ss = s.split(" ");
        int n = pattern.length();
        if (n != ss.length) {
            return false;
        }
        for (int i = 0; i < n; ++i) {
            char ch = pattern.charAt(i);
            if (ch2str.containsKey(ch) && !ch2str.get(ch).equals(ss[i])) {
                return false;
            }
            if (str2ch.containsKey(ss[i]) && !str2ch.get(ss[i]).equals(ch)) {
                return false;
            }
            ch2str.put(ch, ss[i]);
            str2ch.put(ss[i], ch);
        }
        return true;
    }
}

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