Leetcode: Word Pattern II

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:
pattern = "abab", str = "redblueredblue" should return true.
pattern = "aaaa", str = "asdasdasdasd" should return true.
pattern = "aabb", str = "xyzabcxzyabc" should return false.
Notes:
You may assume both pattern and str contains only lowercase letters.

因为目标字符串可以任意划分，所以我们不得不尝试所有可能性。这里通过深度优先搜索的回溯法，对于pattern中每个字母，在str中尝试所有的划分方式，如果划分出来的子串可以用这个字母映射，或者可以建立一个新的字母和字符串的映射关系，我们就继续递归判断下一个pattern中的字母，直到pattern的指针和str的指针同时指到最末

技巧在于，

1. 用HashMap + HashSet来保证一一对应

2. 把HashMap和HashSet用作instance variable, 甚至boolean result也可用作instance variable, 这样任何在递归调用函数里所做的改变，都会保留，而不用把HashMap, HashSet, result加入递归argument

public class Solution {
HashMap<Character, String> map = new HashMap<Character, String>();
HashSet<String> set = new HashSet<String>();

public boolean wordPatternMatch(String pattern, String str) {
if (pattern==null || str==null) return false;
return helper(pattern, str, 0, 0);
}

public boolean helper(String pattern, String str, int i, int j) {
if (i==pattern.length() && j==str.length()) {
return true;
}
if (i==pattern.length() || j==str.length()) return false;
char key = pattern.charAt(i);
for (int cut=j+1; cut<=str.length(); cut++) {
String trial = str.substring(j, cut);
if (!map.containsKey(key) && !set.contains(trial)) { // ensure one-on-one matching
map.put(key, trial);
set.add(trial);
if (helper(pattern, str, i+1, cut))
return true;
map.remove(key);
set.remove(trial);
}
else if (map.containsKey(key) && map.get(key).equals(trial)) {
if (helper(pattern, str, i+1, cut))
return true;
}
}
return false;
}
}