单词搜索

单词搜索

​ 给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

​ 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

题解

​ dfs 深度搜索，逐个寻找单词，思路不难，代码优化的思路比较重要

class Solution {
    int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    public boolean exist(char[][] board, String word) {
        int h = board.length, w = board[0].length;
        boolean[][] visited = new boolean[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                boolean flag = dfs(board, visited, word, 0, i, j);
                if (flag) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean dfs(char[][] board, boolean[][] visited, String word, int index, int x, int y) {
        if(board[x][y] != word.charAt(index)){
            return false;
        }else if(index == word.length() - 1){
            return true;
        }
        visited[x][y] = true;

        for(int[] dir: directions) {
            int next_x = x + dir[0];
            int next_y = y + dir[1];
            if(next_x >= 0 && next_x < board.length && next_y >= 0 && next_y < board[0].length){
                if(!visited[next_x][next_y]){
                    if (dfs(board,visited,word,index + 1,next_x,next_y)) {
                        return true;
                    }
                }
            }
        }
        visited[x][y] = false;
        return false;
    }
}

var directions = [][]int{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}

func exist(board [][]byte, word string) bool {
	h, w := len(board), len(board[0])
	var visited = make([][]bool, h)
	for i := range visited {
		visited[i] = make([]bool, w)
	}
	for i, row := range board {
		for j := range row {
			flag := dfs(board, visited, word, 0, i, j)
			if flag {
				return true
			}
		}
	}
	return false
}

func dfs(board [][]byte, visited [][]bool, word string, index int, x, y int) bool {
	if board[x][y] != word[index] {
		return false
	}
	if index == len(word)-1 {
		return true
	}
	visited[x][y] = true
	for _, dir := range directions {
		next_x, next_y := x+dir[0], y+dir[1]
		if next_x >= 0 && next_x < len(board) && next_y >= 0 && next_y < len(board[0]) && !visited[next_x][next_y] {
			flag := dfs(board, visited, word, index+1, next_x, next_y)
			if flag {
				return true
			}
		}
	}
	visited[x][y] = false
	return false

}