205. Isomorphic Strings

問題描述：

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.

解析：

這個就是判斷兩個字串是否是同構的，最簡單的思路就是構造字元與字元之間的對映關係。

public boolean isIsomorphic(String s, String t) {
       if(s==null || s.length()<=1)
			return true;
		HashMap<Character,Character> hm=new HashMap<Character,Character>();
		for(int i=0;i<s.length();i++){
			char ch1=s.charAt(i);
			char ch2=t.charAt(i);
			if(hm.containsKey(ch1)){
				if(hm.get(ch1)==ch2)
					continue;
				else
					return false;
			}else{
				if(hm.containsValue(ch2))
					return false;
				else
					hm.put(ch1, ch2);
			}
		}
		return true;
    }
 

290. Word Pattern

問題描述：

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:

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.

解析：

這個題目是不是跟上一個很相似，解題思路也是很相似，不過參考了網友的厲害的解題方法，核心程式碼：

//my key
public static boolean wordPattern_My(String pattern, String str) {
		if(pattern==null || str==null)
			return true;
        String[] strs=str.split(" ");
        if(pattern.length()!=strs.length)
            return false;
        HashMap<Character,String> map=new HashMap<Character,String>(); 
        for(int i=0;i<pattern.length();i++){
        	if(map.containsKey(pattern.charAt(i))){
        		if(map.get(pattern.charAt(i)).equals(strs[i]))
        			continue;
        		else
        			return false;
        	}
        	else{
        		if(map.containsValue(strs[i]))
        			return false;
        		else
        			map.put(pattern.charAt(i), strs[i]);
        	}
        }
        return true;
    }
	
	//網友的解析
	public static boolean wordPattern(String pattern, String str) {
		if(pattern==null || str==null)
			return true;
        String[] strs=str.split(" ");
        if(pattern.length()!=strs.length)
            return false;
        HashMap<Object,Integer> map=new HashMap<Object,Integer>(); 
        for(int i=0;i<pattern.length();i++){
        	if(!Objects.equals(map.put(pattern.charAt(i),i),map.put(strs[i],i))){
        		return false;
        	}
        }
        return true;
    }