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。