211. Add and Search Word - Data structure design

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters 

a-z

 or 

.

.
A 

.

 means it can represent any one letter.

For example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

Note:

You may assume that all words are consist of lowercase letters 

a-z

.

就是棵树……不是二叉的 分叉有点多……完了

public class WordDictionary {

class TrieNode {
boolean endHere;  // empty string, 1 node; string of length n, (n + 1) nodes.
TrieNode[] branch;

TrieNode () {
endHere = false;
branch = new TrieNode[26];
}
}

TrieNode root = new TrieNode();

// Adds a word into the data structure.
public void addWord(String word) {
addWord(word, 0, root);
}

private void addWord(String word, int p, TrieNode node) {
if (p == word.length()) {
node.endHere = true;
return;
}
int index = word.charAt(p) - 'a';
TrieNode succ = node.branch[index];
if (succ == null) {
succ = new TrieNode();
node.branch[index] = succ;
}

4000
addWord(word, p + 1, succ);
}

// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
public boolean search(String word) {
return search(word, 0, root);
}

private boolean search(String word, int p, TrieNode node) {
if (p == word.length()) { return node.endHere; }
char c = word.charAt(p);
if (c != '.') {
if (node.branch[c - 'a'] == null) { return false; }
return search(word, p + 1, node.branch[c - 'a']);
} else {
for (int i = 0; i < 26; ++i) {
if (node.branch[i] != null && search(word, p + 1, node.branch[i])) {
return true;
}
}
}
return false;
}
}