leetcode [#290]

目录

• 1. 题目
• 2. 解决方案
• 3. 注意事项

题目

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. pattern = “abba”, str = “dog cat cat dog” should return true.
2. pattern = “abba”, str = “dog cat cat fish” should return false.
3. pattern = “aaaa”, str = “dog cat cat dog” should return false.
4. pattern = “abba”, str = “dog dog dog dog” should return false.

Note:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

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解决方案

import java.util.Hashtable;
public class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] arr = str.split(" ");
        int len1 = pattern.length();
        int len2 = arr.length;
        if(len1 != len2) return false;

        Hashtable<Character, String> table = new Hashtable<>();
        for(int i = 0; i < len1; i++){
            if(!table.containsKey(pattern.charAt(i))){
                if(!table.containsValue(arr[i])){
                    table.put(pattern.charAt(i), arr[i]);
                } else {
                    return false;
                }
            } else {
                if(!table.get(pattern.charAt(i)).equals(arr[i])){
                    return false;
                }
            }
        }
        return true;
    }
}

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注意事项

1.使用一个Hashtable，key为模式字符串，value为待检测字符串。遍历两个输入字符串，如果key值pattern.charAt(i)不存在，且arr[i]的值在table中不存在，则可以将pattern.charAt(i)和arr[i]对应起来，存入table。

1. 如果key值pattern.charAt(i)存在，则将其value值和当前arr[i]比较，如不同，说明模式未能正确匹配，返回false。