LeetCode刷题笔录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.

这题好像除了用DFS暴力搜索没什么别的好办法。

用两层循环，对每一个字符进行一个DFS，而且可以往4个方向走。

另外用一个boolean[][] visited来记录当前字符是否被visit过。注意在递归后要重置visited。

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

	public boolean recursive(char[][] board, String word, int index, int row,
			int col, int rowLimit, int colLimit, boolean[][] visited) {
		if (index == word.length())
			return true;
		// out of bounds
		if (row < 0 || col < 0 || row >= rowLimit || col >= colLimit)
			return false;
		// at a visited node
		if (visited[row][col])
			return false;
		// the two chars do not match
		if (board[row][col] != word.charAt(index))
			return false;
		// dfs
		visited[row][col] = true;
		boolean result = recursive(board, word, index + 1, row + 1, col,
				rowLimit, colLimit, visited)
				|| recursive(board, word, index + 1, row - 1, col, rowLimit,
						colLimit, visited)
				|| recursive(board, word, index + 1, row, col + 1, rowLimit,
						colLimit, visited)
				|| recursive(board, word, index + 1, row, col - 1, rowLimit,
						colLimit, visited);
		visited[row][col] = false;
		return result;
	}
}