Word Search

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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

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

public class Solution {
    public boolean exist(char[][] board, String word) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if(word.length() < 1)
			return false;
		int row = board.length;
		if(row == 0)
			return false;
		int col = board[0].length;
		if(col == 0)
			return false;
		for(int i = 0; i < row; i++){
			for(int j = 0; j < col; j++){
				char cur = word.charAt(0);
				if(board[i][j] == cur){
					board[i][j] = '$';
					if(find(board, word, 1, i, j))
						return true;
					board[i][j] = cur;
				}
			}
		}
		return false;
	}
	public boolean find(char[][] board, String word, int index, int row, int col){
		if(index == word.length())
			return true;
		int nRow = board.length;
		int nCol = board[0].length;
		char target = word.charAt(index);
		if(row - 1 >= 0 && board[row - 1][col] == target){
			board[row - 1][col] = '$';
			if(find(board, word, index + 1, row - 1, col))
				return true;
			board[row - 1][col] = target;
		}
		if(col + 1 < nCol && board[row][col + 1] == target){
			board[row][col + 1] = '$';
			if(find(board, word, index + 1, row, col + 1))
				return true;
			board[row][col + 1] = target;
		}
		if(row + 1 < nRow && board[row + 1][col] == target){
			board[row + 1][col] = '$';
			if(find(board, word, index + 1, row + 1, col))
				return true;
			board[row + 1][col] = target;
		}
		if(col - 1 >= 0 && board[row][col - 1] == target){
			board[row][col - 1] = '$';
			if(find(board, word, index + 1, row, col - 1))
				return true;
			board[row][col - 1] = target;
		}
		return false;
	}
}