leetcode 290:Word Pattern

題目描述：

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.

解題思路：

第一種方案：

使用一個hashMap《Character,String》 key保存pattern中的字符，value保存words中的字符串，
比較相同key值的value是否相同

public  boolean wordPattern(String pattern , String str){
        HashMap<Character,String> map = new HashMap<Character,String>();
        String[] words = str.split(" ");
        if(words.length!=pattern.length())
            return false;
        for(int i =0 ; i<words.length;i++){
            if(!map.containsKey(pattern.charAt(i))){
                if(map.containsValue(words[i]))
                    return false;
                map.put(pattern.charAt(i), words[i]);
            }else if(!map.get(pattern.charAt(i)).equals(words[i])){
                return false;
            }
        }
        return true;
    }

第二種方案：
利用hashMap put方法返回值的特性，比較pattern 和word對應的下標是否一致

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

    }