leetcode 290:Word Pattern

题目描述：

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.

解题思路：

第一种方案：

使用一个hashMap《Character,String》 key保存pattern中的字符，value保存words中的字符串，
比较相同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;    }

第二种方案：
利用hashMap put方法返回值的特性，比较pattern 和word对应的下标是否一致

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;    }