Leetcode079-wordSearch

Description

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.

Example

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

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

Solution

class Solution {
    static private int[][] dirt = new int[][]{{1, 0},{-1, 0},{0, 1},{0, -1}};
    public boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0 || board[0].length == 0) return false;
        if (word == null || word.length() == 0) return true;
        char[] w = word.toCharArray();
        boolean[][] visited = new boolean[board.length][board[0].length];
        for (int i = 0; i < board.length; i++)
            for (int j = 0; j < board[0].length; j++)
                if (dfs(board, i, j, w, 0, visited)) return true;                            
        return false;
    }
    public boolean dfs(char[][] board, int x, int y, char[] w, int cur, boolean[][] visited){
        if (board[x][y] != w[cur]) return false;
        if (cur == w.length - 1) return true;
        visited[x][y] = true;
        for (int i = 0; i < 4; i++){
            int nx = x + dirt[i][0];
            int ny = y + dirt[i][1];
            if (nx >= 0 && nx < board.length && ny >=0 && ny < board[0].length && !visited[nx][ny]){
                if (dfs(board, nx, ny, w, cur + 1, visited)) return true;
            }
        }
        visited[x][y] = false;
        return false;
    }
}

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