Java for Leetcode 212 Word Search II

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

Each word must is constructed from letters of sequentially adjacent cell, where "adjacent" cells is those horizontally or Vertically neighboring. The same letter cell is 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"] .

Problem Solving Ideas:

Subject to Java for Leetcode 079 word Search But with DFS, the real practice is to put Word into trie, and then iterate through each point in the trie to see if it matches the elements in the trie, the Java implementation is as follows:

public class Solution {public list<string> findwords (char[][] board, string[] words) {hashset<string> list=n EW HashSet ();          Trie Trie = new Trie ();           for (String word:words) Trie.insert (word);        Boolean[][] Visited=new boolean[board.length][board[0].length]; for (int i = 0, i < board.length; i++) for (int j = 0; J < Board[0].length; j + +) DFS (list           , board, visited, "", I, J, Trie);      return new ArrayList (list); } public void Dfs (set<string> list,char[][] board, boolean[][] visited, String str, int x, int y, Trie tr          IE) {if (x < 0 | | x >= board.length | | y < 0 | | y >= board[0].length) return;          if (Visited[x][y]) return;          str + = Board[x][y];          if (!trie.startswith (str)) return;        if (Trie.search (str)) List.add (str);          Visited[x][y] = true; DFS (List,board, visited, str, x-1, y, trie);          DFS (List,board, visited, str, x + 1, y, trie);          DFS (List,board, visited, str, x, y-1, trie);          DFS (List,board, visited, str, x, y + 1, trie);      Visited[x][y] = false; }}class Trienode {//Initialize your data structure Here.int num;//How many words pass through this node, that is, the number of occurrences of the node character trienode[] son;//all son nodes Boole An isend;//is not the last node of the Char val;//node value Trienode () {this.num = 1;this.son = new Trienode[26];this.isend = false;}} Class Trie {protected Trienode root;public Trie () {root = new Trienode ();} public void Insert (String word) {if (Word = = NULL | | word.length () = = 0) return; Trienode node = this.root;char[] Letters = Word.tochararray (); for (int i = 0; i < word.length (); i++) {Int. POS = Letter S[i]-' a ', if (node.son[pos] = = null) {Node.son[pos] = new Trienode (); node.son[pos].val = Letters[i];} else {Node.son[pos] . num++;} node = Node.son[pos];} Node.isend = true;} Returns If the word is in the Trie.public boolean search (String word) {if (Word = = NULL | | word.length () = = 0) {return false;} Trienode node = root;char[] Letters = Word.tochararray (); for (int i = 0; i < word.length (); i++) {int pos = letters[i] -' a '; if (node.son[pos]! = null) {node = Node.son[pos];} else {return false;}} return node.isend;} Returns If there is any word in the trie//this starts with the given Prefix.public boolean startsWith (String prefix) { if (prefix = = NULL | | prefix.length () = = 0) {return false;} Trienode node = root;char[] Letters = Prefix.tochararray (); for (int i = 0; i < prefix.length (); i++) {int pos = letters [i]-' a '; if (node.son[pos]! = null) {node = Node.son[pos];} else {return false;}} return true;}}

Java for Leetcode 212 Word Search II