单词搜索 | 徐年の博客

力扣 79. 单词搜索

题目说明

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。

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

示例

示例 1:

1 2	输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" 输出：true

示例 2:

1 2	输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" 输出：true

示例 3:

1 2	输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" 输出：false

笔者理解

此题是一道字符串算法问题，在力扣题库中被定义为中等题。

解法

当笔者阅读完此题后，发现我们直接DFS进行判断即可，让我们来看看具体如何实现的吧。

实现

class Solution {
    /**
     * DFS
     */

    // 转向
    int[] turnX = {-1, 1, 0, 0};
    int[] turnY = {0, 0, -1, 1};

    int m, n;

    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();

        m = board.length;
        n = board[0].length;

        boolean[][] ticks = new boolean[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == words[0]) {
                    ticks[i][j] = true;
                    if (dfs(board, words, 1, i, j, ticks)) {
                        return true;
                    }
                    ticks[i][j] = false;
                }
            }
        }

        return false;
    }

    /**
     * DFS
     */
    public boolean dfs(char[][] board, char[] words, int now, int i, int j, boolean[][] ticks) {
        // 已经匹配完成
        if (now == words.length) {
            return true;
        }

        boolean result = false;

        for (int k = 0; k < 4; k++) {
            i += turnX[k]; j += turnY[k];

            // 不符合要求排除
            if (!judge(i, j) && !ticks[i][j] && board[i][j] == words[now]) {
                ticks[i][j] = true;
                result |= dfs(board, words, now + 1, i, j, ticks);
                ticks[i][j] = false;
            }

            i -= turnX[k]; j -= turnY[k];
        }

        return result;
    }

    /**
     * 判断下标是否越界
     */
    public boolean judge(int i, int j) {
        return i < 0 || j < 0 || (i - m) * (j - n) <= 0;
    }
}