290. Word Pattern

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。