LeetCode Word Pattern II

原题链接在这里：https://leetcode.com/problems/word-pattern-ii/

题目：

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 substring in str.

Examples:

1. pattern = "abab", str = "redblueredblue" should return true.
2. pattern = "aaaa", str = "asdasdasdasd" should return true.
3. pattern = "aabb", str = "xyzabcxzyabc" should return false.

Notes:
You may assume both pattern and str contains only lowercase letters.

题解：

类似Word Pattern. 但这里没有明确分割words的空格.

采用backtracking, 用pattern内的一个char来试着match不同长度的str substring, 能同时试到pattern 和 str的结尾, 就返回true.

e.g. pattern = "abab", str = "redblueredblue", 先用"a" match "r", "b" match "e", 又遇到"a", 但str里确实"d", 不match.

就backtracking下然后用"b" match "ed", 以此类推.

用HashMap<Character, String> 来保存已经走过的pattern char 和 str substring对应关系.

同时maintain一个set用来避免相同的string 来match 不同的char. e.g. "a" match "cde", "b"也match "cde".

Time Complexity: exponential. Space: O(hm.size() + hs.size()).

AC Java: 

public class Solution {
    public boolean wordPatternMatch(String pattern, String str) {
        HashMap<Character, String> hm = new HashMap<Character, String>();
        HashSet<String> hs = new HashSet<String>();
        return isMatch(pattern, 0, str, 0, hm, hs);
    }
    
    private boolean isMatch(String pattern, int i, String str, int j, HashMap<Character, String> hm, HashSet<String> hs){
        //stop condition
        if(i == pattern.length() && j == str.length()){
            return true;
        }
        if(i == pattern.length() || j == str.length()){
            return false;
        }
        
        char c = pattern.charAt(i);
        
        //If current pattern character already exists in map
        if(hm.containsKey(c)){
            String cString = hm.get(c);
            
            //If corresponding string does not match str.substring(j, j+matchedString.length()).
            if(!str.startsWith(cString, j)){
                return false;
            }
            return isMatch(pattern, i+1, str, j+cString.length(), hm, hs);
        }else{
            //current pattern character doesn't exist in the map
            for(int k = j; k<str.length(); k++){
                String s = str.substring(j, k+1);
                
                //different strings can't be matched to same char
                if(hs.contains(s)){
                    continue;
                }
                
                hm.put(c, s);
                hs.add(s);
                
                //if all reset string could be matched
                if(isMatch(pattern, i+1, str, k+1, hm, hs)){
                    return true;
                }
                
                hm.remove(c);
                hs.remove(s);
            }
            
            //Have tried all different lengths
            return false;
        }
    }
}