leetcode JAVA Word Search 难度系数3 3.28

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 =

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

public class Solution {
    private int row;
	private int col;
    public boolean exist(char[][] board, String word) {
		row = board.length;
		col = board[0].length;
		boolean[][] visited = new boolean[row][col];
		for (int i = 0; i < row; i++) {
			for (int j = 0; j < col; j++) {
				if (dfs(board, word, 0, i, j, visited))
					return true;
			}
		}
		return false;
	}

	private boolean dfs(char[][] board, String word, int index, int rowindex,
			int colindex, boolean[][] visited) {
		if (index == word.length())
			return true;
		if (rowindex < 0 || colindex < 0 || rowindex >= row || colindex >= col)
			return false;
		if (visited[rowindex][colindex])
			return false;
		if (board[rowindex][colindex] != word.charAt(index))
			return false;
		visited[rowindex][colindex] = true;
		boolean res = dfs(board, word, index + 1, rowindex - 1, colindex,
				visited)
				|| dfs(board, word, index + 1, rowindex + 1, colindex, visited)
				|| dfs(board, word, index + 1, rowindex, colindex + 1, visited)
				|| dfs(board, word, index + 1, rowindex, colindex - 1, visited);
		visited[rowindex][colindex] = false;
		return res;

leetcode JAVA Word Search 难度系数3 3.28

原文：http://blog.csdn.net/yiding_he/article/details/18893621