290. Word Pattern

叁歲伎倆 2022-07-15 01:29 67阅读 0赞

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`.

Examples:

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.

Notes:

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

解法一：

import java.util.Hashtable;
    public class Solution {
        char index = 'a';
        public boolean wordPattern(String pattern, String str) {
            if (pattern.length() != str.split(" ").length)
    			return false;
    		Map<String, Character> table = new Hashtable<String, Character>();
    		StringBuilder sb = new StringBuilder();
    		for(int i=0;i<pattern.length();i++){
    			if(table.get(String.valueOf(pattern.charAt(i)))==null){
    				table.put(String.valueOf(pattern.charAt(i)), index++);
    			}
    			sb.append(table.get(String.valueOf(pattern.charAt(i))));
    		}
    		String[] sub = str.split(" ");
    		index = 'a';
    		table.clear();
    		for (int i = 0; i < sub.length; i++) {
    			if (table.get(sub[i]) == null)
    				table.put(sub[i], index++);
    			if (table.get(sub[i]) != sb.charAt(i))
    				return false;
    		}
    		return true;
        }
    }

解法二：

public class Solution {
    public boolean wordPattern(String pattern, String str) {
    		String[] words = str.split(" ");
    		if (words.length != pattern.length())
    			return false;
    		Map index = new HashMap();
    		for (Integer i = 0; i < words.length; ++i)
    			if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
    				return false;
    		return true;
    	}
    }

注：HashMap中put（Key k，Value v）方法的返回值：若键为k的键值对不存在，则返回null；否则返回put执行之前map中键为k的键值对的Value值。