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<Character, String> map = new HashMap<>();
return helper(pattern, 0, str, 0, map);
}
private boolean helper(String pattern, int i, String str, int j, Map<Character, String> 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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