Word Pattern

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.

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

reverseMap 可以用set来代替。

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] strArray = str.split(" ");
        if(pattern.length() != strArray.length)
            return false;
        
        Map<Character, String> map = new HashMap<>();
        Set<String> reverseRecord = new HashSet<>();
        for(int i = 0; i < strArray.length; i++){
            Character currentChar = pattern.charAt(i);
            String currentString = strArray[i];
            if(!map.containsKey(currentChar)){
                //when we reach this step, we know that currentChar has never appeared before, so we expect that currentString
                //has not appeated either. If it has appeared, then it must not map to currentChar
                if(reverseRecord.contains(currentString))
                    return false;
                else{
                    map.put(currentChar, currentString);
                    reverseRecord.add(currentString);
                }
            }
            else{
                if( !map.get(currentChar).equals(currentString))
                    return false;
            }
        }
        return true;
    }
}