zoukankan      html  css  js  c++  java
  • 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 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;
        }
    }
  • 相关阅读:
    jquery绑定点击事件动画BUG
    初步了解XSS攻击
    构造函数、原型对象、原型链之间的关系
    SQA计划和系统测试规程
    第四次scrum冲刺
    第二次Scrum冲刺
    前端面试题整理
    vue 2 简化版数据响应原理
    Vue3.0 简化版数据响应式原理
    git commit规范
  • 原文地址:https://www.cnblogs.com/renyualbert/p/5894746.html
Copyright © 2011-2022 走看看