一、题目粗心

标签: 搜寻

https://leetcode.cn/problems/word-search

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 。

单词必须依照字母程序，通过相邻的单元格内的字母形成，其中“相邻”单元格是那些程度相邻或垂直相邻的单元格。同一个单元格内的字母不容许被重复使用。

示例 1：

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

示例 2：

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

示例 3：

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

提醒：

  • 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

进阶：你能够应用搜寻剪枝的技术来优化解决方案，使其在 board 更大的状况下能够更快解决问题？

二、解题思路

还是用回溯法，定义一个二维数组存储拜访标记，在对任意地位进行深度优先搜寻时，先将以后地位为已拜访，以防止反复遍历，在所有的可能都搜寻实现后，再改回以后地位为未拜访，避免烦扰其它地位搜寻以后地位。

三、解题办法

3.1 Java实现

public class Solution {    boolean find = false;    public boolean exist(char[][] board, String word) {        if (board.length == 0) {            return false;        }        int m = board.length;        int n = board[0].length;        boolean[][] visited = new boolean[m][n];        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                backTracking(i, j, board, word, visited, 0);            }        }        return find;    }    void backTracking(int i, int j, char[][] board, String word, boolean[][] visited, int pos) {        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length) {            return;        }        if (visited[i][j] || find || board[i][j] != word.charAt(pos)) {            return;        }        if (pos == word.length() - 1) {            find = true;            return;        }        visited[i][j] = true;        backTracking(i + 1, j, board, word, visited, pos + 1);        backTracking(i - 1, j, board, word, visited, pos + 1);        backTracking(i, j + 1, board, word, visited, pos + 1);        backTracking(i, j - 1, board, word, visited, pos + 1);        visited[i][j] = false;    }}

四、总结小记

  • 2021/6/5 明天假日最初一天