LeetCode Word Pattern II

Posted 2020-08-29 Dylan_Java_NYC

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: 

 1 public class Solution {
 2     public boolean wordPatternMatch(String pattern, String str) {
 3         HashMap<Character, String> hm = new HashMap<Character, String>();
 4         HashSet<String> hs = new HashSet<String>();
 5         return isMatch(pattern, 0, str, 0, hm, hs);
 6     }
 7     
 8     private boolean isMatch(String pattern, int i, String str, int j, HashMap<Character, String> hm, HashSet<String> hs){
 9         //stop condition
10         if(i == pattern.length() && j == str.length()){
11             return true;
12         }
13         if(i == pattern.length() || j == str.length()){
14             return false;
15         }
16         
17         char c = pattern.charAt(i);
18         
19         //If current pattern character already exists in map
20         if(hm.containsKey(c)){
21             String cString = hm.get(c);
22             
23             //If corresponding string does not match str.substring(j, j+matchedString.length()).
24             if(!str.startsWith(cString, j)){
25                 return false;
26             }
27             return isMatch(pattern, i+1, str, j+cString.length(), hm, hs);
28         }else{
29             //current pattern character doesn\'t exist in the map
30             for(int k = j; k<str.length(); k++){
31                 String s = str.substring(j, k+1);
32                 
33                 //different strings can\'t be matched to same char
34                 if(hs.contains(s)){
35                     continue;
36                 }
37                 
38                 hm.put(c, s);
39                 hs.add(s);
40                 
41                 //if all reset string could be matched
42                 if(isMatch(pattern, i+1, str, k+1, hm, hs)){
43                     return true;
44                 }
45                 
46                 hm.remove(c);
47                 hs.remove(s);
48             }
49             
50             //Have tried all different lengths
51             return false;
52         }
53     }
54 }