Word Search II

题目：

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

For example,

Given words =

["oath","pea","eat","rain"]

and board =

[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]

Return

["eat","oath"]

.

Note:

You may assume that all inputs are consist of lowercase letters

a-z

.
分析：
借助字典树

参考代码：
http://www.programcreek.com/2014/06/leetcode-word-search-ii-java/

public class Solution {
Set<String> result = new HashSet<String>();

public List<String> findWords(char[][] board, String[] words) {
//HashSet<String> result = new HashSet<String>();

Trie trie = new Trie();
for(String word: words){
trie.insert(word);
}

int m=board.length;
int n=board[0].length;

boolean[][] visited = new boolean[m]
;

for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
dfs(board, visited, "", i, j, trie);
}
}

return new ArrayList<String>(result);
}

public void dfs(char[][] board, boolean[][] visited, String str, int i, int j, Trie trie){
int m=board.length;
int n=board[0].length;

if(i<0 || j<0||i>=m||j>=n){
return;
}

if(visited[i][j])
return;

str = str + board[i][j];

if(!trie.startsWith(str))
return;

if(trie.search(str)){
result.add(str);
}

visited[i][j]=true;
dfs(board, visited, str, i-1, j, trie);
dfs(board, visited, str, i+1, j, trie);
dfs(board, visited, str, i, j-1, trie);
dfs(board, visited, str, i, j+1, trie);
visited[i][j]=false;
}
}
//Trie Node
class TrieNode{
public TrieNode[] children = new TrieNode[26];
public String item = "";
}

//Trie
class Trie{
public TrieNode root = new TrieNode();

public void insert(String word){
TrieNode node = root;
for(char c: word.toCharArray()){
if(node.children[c-'a']==null){
node.children[c-'a']= new TrieNode();
}
node = node.children[c-'a'];
}
node.item = word;
}

public boolean search(String word){
TrieNode node = root;
for(char c: word.toCharArray()){
if(node.children[c-'a']==null)
return false;
node = node.children[c-'a'];
}
if(node.item.equals(word)){
return true;
}else{
return false;
}
}

public boolean startsWith(String prefix){
TrieNode node = root;
for(char c: prefix.toCharArray()){
if(node.children[c-'a']==null)
return false;
node = node.children[c-'a'];
}
return true;
}
}