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.
public class Solution { public IList<string> FindWords(char[,] board, string[] words) { var res = new List<string>(); if(words.Count() == 0 || board == null) return res; Trie trie = new Trie(); foreach(string word in words) { trie.Insert(word); } var visited = new bool[board.GetLength(0),board.GetLength(1)]; for(int i = 0;i<board.GetLength(0) ;i++) { for(int j = 0;j<board.GetLength(1) ;j++) { if(trie.Root.Child[board[i,j]-‘a‘] != null) { Search(board,trie.Root.Child[board[i,j]-‘a‘], visited,i,j,res,board[i,j]+""); } } } return res; } private void Search(char[,] board,TrieNode t, bool[,] visited, int row, int col,IList<string> res, string cur) { if(t != null && t.EndWord) { if(!res.Contains(cur))res.Add(cur); } if(row<0 || row >= board.GetLength(0)|| col<0 || col >= board.GetLength(1)) { return; } else { visited[row,col] = true; if(row>0 && !visited[row-1,col] && t.Child[board[row-1,col]-‘a‘] != null ) Search(board,t.Child[board[row-1,col]-‘a‘], visited, row-1, col,res, cur+board[row-1,col]); if(row<board.GetLength(0)-1 && !visited[row+1,col] && t.Child[board[row+1,col]-‘a‘] != null) Search(board,t.Child[board[row+1,col]-‘a‘], visited, row+1, col,res, cur+board[row+1,col]); if(col>0 &&!visited[row,col-1] && t.Child[board[row,col-1]-‘a‘] != null ) Search(board,t.Child[board[row,col-1]-‘a‘], visited, row, col-1,res, cur+board[row,col-1]); if(col<board.GetLength(1)-1 && !visited[row,col+1] && t.Child[board[row,col+1]-‘a‘] != null) Search(board,t.Child[board[row,col+1]-‘a‘], visited, row, col+1,res, cur+board[row,col+1]); visited[row,col] = false; } } } public class TrieNode { public TrieNode[] Child {get;set;} public bool EndWord {get;set;} public TrieNode() { Child = new TrieNode[26]; EndWord = false; } } public class Trie{ public TrieNode Root {get;set;} public Trie() { Root = new TrieNode(); } public void Insert(string word) { var sentinel = Root; foreach(char c in word) { if(sentinel.Child[c-‘a‘] == null) sentinel.Child[c-‘a‘] = new TrieNode(); sentinel = sentinel.Child[c-‘a‘]; } sentinel.EndWord = true; } }
原文：http://www.cnblogs.com/renyualbert/p/5894746.html