Word Search

1. Title

Word Search

2.   Http address

https://leetcode.com/problems/word-search/

3. The question

Given a 2D board and a word, find if the word exists in the grid.

The word can 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.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

4. My code (AC)

 1 public class WordSearch {
 2     
 3     public static void main(String[] args) {
 4         
 5     }
 6 
 7     public boolean DFS(char[][] board,int i, int j, String word,  boolean [][]visited) {
 8         
 9         if( word == null || word.equals("") )
10                 return true;
11 
12         if( word.length() == 1 && word.charAt(0) == board[i][j] )
13         {
14         //        System.out.println(true);    
15                 return true;
16         }
17         if( word.length() == 1 && word.charAt(0) != board[i][j] )
18                 return false;
19         int x = 0 , y = 0;
20         String subStr = word.substring(1);
21 
22         visited[i][j] = true;
23 
24         if( board[i][j] != word.charAt(0) )
25         {
26             visited[i][j] = false;
27             return false;
28         }
29 
30         //up
31         x = i - 1;
32         y = j;
33         if( x >= 0  && visited[x][y] == false)
34         {
35             if ( DFS(board, x, y, word.substring(1), visited) )
36                     return true;
37         }
38         //down
39         x = i + 1;
40         y = j;
41         if( x  < board.length && visited[x][y] == false)
42         {
43             if ( DFS(board, x, y, word.substring(1), visited) )
44                     return true;
45         }
46         //left
47         x = i;
48         y = j - 1;
49         if( y >= 0  && visited[x][y] == false)
50         {
51             if ( DFS(board, x, y, word.substring(1), visited) )
52                 return true;
53         }
54         //right
55         x = i;
56         y = j + 1;
57         if( y < board[0].length && visited[x][y] == false)
58         {
59             if ( DFS(board, x, y, word.substring(1), visited) )
60                 return true;
61         }
62         visited[i][j] = false;
63         return false;
64     }
65     // Accepted
66       public boolean exist(char[][] board, String word) {
67          
68           boolean [][]visited = new boolean[board.length][board[0].length];
69             for(int i = 0 ; i < board.length; i++ )
70             {
71 
72                 for(int j = 0 ; j < board[0].length; j++ )
73                 {
74                     if(  DFS(board, i , j, word, visited) )
75                             return true;
76                 }
77             }
78             return false;
79         }
80 }

 

转载于:https://www.cnblogs.com/ordili/p/4928287.html