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 }