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;
    }
}