LeetCode Word Search

题目：

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.

class Solution {
public:
	bool exist(vector<vector<char> > &board, string word) {
		if (word.size() == 0)
			return true;
		m = board.size();
		n = board[0].size();
		//上下左右扩展
		step[0][0] = -1;
		step[0][1] = 0;
		step[1][0] = 1;
		step[1][1] = 0;
		step[2][0] = 0;
		step[2][1] = -1;
		step[3][0] = 0;
		step[3][1] = 1;
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++)
				used[i][j] = false;
		}
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if (board[i][j] == word[0]) {
					used[i][j] = true;
					flag = false;
					dfs(1, word.size(),i, j, board, word);
					if (flag)
						return true;
					used[i][j] = false;
				}
			}
		}
		return false;
	}
private:
	bool used[1000][1000];
	bool flag;
	int step[4][2];
	int m, n;

	void dfs(int depth, int maxDepth, int x, int y, vector<vector<char> > &board, string &word) {
	    if(flag)
	        return;
		if (depth == maxDepth) {
			flag = true;
			return;
		}
		for (int i = 0; i < 4; i++) {
			int nx = x + step[i][0];
			int ny = y + step[i][1];
			if (0 <= nx && nx < m && 0 <= ny && ny < n && !used[nx][ny] && board[nx][ny] == word[depth]) {
				used[nx][ny] = true;
				dfs(depth + 1, maxDepth, nx, ny, board, word);
				used[nx][ny] = false;
			}
		}
	}
};