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  • 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 }
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  • 原文地址:https://www.cnblogs.com/ordili/p/4928287.html
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