Word Pattern II 解答

Question

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.

Solution

Word Pattern 是用HashMap的containsKey()和containsValue()两个方法实现的。

但是对于Word Pattern II，我们并不知道怎样划分str，所以基本想法是“暴力搜索”即backtracking/DFS。这里DFS的存储对象不是常见的list，而是map.

 public class Solution {
     public boolean wordPatternMatch(String pattern, String str) {
         return helper(pattern, 0, str, 0, new HashMap<Character, String>());
     }

     private boolean helper(String pattern, int startP, String str, int startS, Map<Character, String> map) {
         if (startP == pattern.length() && startS == str.length()) {
             return true;
         } else if (startP == pattern.length()) {

             }
             if (match.equals(str.substring(startS, endS))) {
                 return helper(pattern, startP + 1, str, endS, map);
             } else {
                 return false;
             }
         } else {
             // If map does not have existing key
             // Traverse, brute force

             for (int i = startS + 1; i <= str.length(); i++) {
                 String candidate = str.substring(startS, i);
                 if (map.containsValue(candidate)) {
                     continue;
                 }
                 map.put(key, candidate);
                 if (helper(pattern, startP + 1, str, i, map)) {
                     return true;
                 }
                 map.remove(key);
             }
         }
         return false;
     }
 }