​package cn.edu.xidian.sselab.hashtable;

import java.util.HashMap;
import java.util.Map;

/**
 *
 * @author zhiyong wang
 * title: Word Pattern
 * content:
 * 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:
 *
 *    pattern = "abba", str = "dog cat cat dog" should return true.
 *    pattern = "abba", str = "dog cat cat fish" should return false.
 *    pattern = "aaaa", str = "dog cat cat dog" should return false.
 *    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.
 *
 */
public class WordPattern {
    
    //这个地方犯了两次错误，怎么就一直不改正呢
    //没有把key不同，value相同的情况排除掉  即没有排除：abab,dog dog dog dog这种情况
    public boolean wordPattern(String pattern, String str){
        if(pattern == null || str == null) return false;
        int lengthP = pattern.length();
        String[] strArray = str.split(" ");
        int lengthS =  strArray.length;
        if(lengthP == 0 || lengthS == 0 || lengthP != lengthS) return false;
        HashMap<Character,String> container =  new HashMap<Character,String>();
        for(int i=0;i<lengthP;i++){
            if(container.containsKey(pattern.charAt(i))){
                if(!container.get(pattern.charAt(i)).equals(strArray[i])){
                    return false;
                }
            }else{
                if(container.containsValue(strArray[i])){
                    return false;
                }else{
                    container.put(pattern.charAt(i), strArray[i]);
                }                
            }
        }    
        return true;
    }
    
    //大牛的算法,这里学习了index.put(key,value)的返回值，根据key来判断
    /*
     * 这是最原始的put的返回值的注释
     * the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>,
     *         if the implementation supports <tt>null</tt> values.)
     *  大意是如果put的key值原来并不在map中，则返回null，如果key已经存在了，那么返回key值对应的前一个value值       
     *  用这种方法判断，key存放的pattern的每一个字符，与str的每一个单词，然后根据对应的位置value来判断是否是一一影射
     *     这里注意一下返回值是一个对象
     * */
    public boolean wordPatterns(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++){//这里一开始用int报错了
            if(index.put(pattern.charAt(i),i) != index.put(words[i],i))
                return false;
        }
        return true;
    }
    
    public static void main(String[] args) {
        WordPattern word = new WordPattern();
        word.wordPatterns("abba", "dog dog cat cat");
    }
    
}