Word Pattern

​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");
}

}