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:

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.

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;
}
}