212. Word Search II(Trie)

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"]

.

查找若干个单词是否在字符矩阵中，可先将若干单词建立Trie树，然后dfs字符矩阵中的单词是否在Trie树中

public class Solution {
Set<String> res = new HashSet<String>();
int[][] dir = {{0,1},{0,-1},{1,0},{-1,0}};
public List<String> findWords(char[][] board, String[] words) {
Trie trie = new Trie();
for(String str:words)
{
trie.insert(str);
}
int n= board.length;
int m = board[0].length;
boolean map[][] = new boolean
[m];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
dfs(i,j,board,"",map,trie);
}
}
return new ArrayList<String>(res);
}
private void dfs(int i,int j,char[][] board,String str,boolean map[][],Trie trie)
{
if(i<0 || i>board.length-1 || j<0 || j>board[0].length-1)return;
if(map[i][j])return;
str +=board[i][j];
if(!trie.startsWith(str))return ;
if(trie.search(str))res.add(str);
map[i][j]=true;
for(int ii=0;ii<4;ii++)
{
dfs(i+dir[ii][0],j+dir[ii][1],board,str,map,trie);
}
map[i][j]=false;
}
}
class TrieNode {
boolean isend;
HashMap<Character, TrieNode> map ;
// Initialize your data structure here.
public TrieNode() {
this.isend = false;
this.map = new HashMap<Character, TrieNode>();
}
}