[LeetCode] 291. Word Pattern II

genefy 发布于2019-08-16 13:44 / 2796人阅读

Problem

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.

Example 1:

Input: pattern = "abab", str = "redblueredblue"
Output: true
Example 2:

Input: pattern = pattern = "aaaa", str = "asdasdasdasd"
Output: true
Example 3:

Input: pattern = "aabb", str = "xyzabcxzyabc"
Output: false
Notes:
You may assume both pattern and str contains only lowercase letters.

Solution

class Solution {
    public boolean wordPatternMatch(String pattern, String str) {
        if (pattern == null || str == null || str.length() < pattern.length()) return false;
        Map map = new HashMap<>();
        return helper(pattern, 0, str, 0, map);
    }
    private boolean helper(String pattern, int i, String str, int j, Map map) {
        if (i == pattern.length() && j == str.length()) return true;
        if (i == pattern.length() || j == str.length()) return false;
        
        char ch = pattern.charAt(i);
        
        if (map.containsKey(ch)) {
            String s = map.get(ch);
            if (!str.substring(j).startsWith(s)) return false;
            else return helper(pattern, i+1, str, j+s.length(), map);
        } else {
            //try put some str substrings into the map
            for (int k = j; k < str.length(); k++) {
                String part = str.substring(j, k+1);
                if (map.containsValue(part)) continue; //already used by a different key
                
                //add the new pair, dfs, remove the new pair (backtracking)
                map.put(ch, part);
                if (helper(pattern, i+1, str, k+1, map)) return true;
                map.remove(ch);
            }
        }
        
        return false;
    }
}

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