算方法学习之二维平面上的回溯法

题目要求：

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
进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/word-search

解题代码：

class Solution {
private:
    int d[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    int m, n;
    vector<vector<bool>> visited;

    bool inArea(int x, int y) {
        return x >= 0 && x < m && y >= 0 && y <n;
    }

    // 从board[startx][starty]开始，寻找word[index...word.size())
    bool searchWord(const vector<vector<char>>& board, const string& word, int index, int startx, int starty) {
        if (index == word.size() - 1)
            return board[startx][starty] == word[index];

        if (board[startx][starty] == word[index]) {
            visited[startx][starty] = true;
            // 从startx,starty出发，向四个方向寻找
            for (int i = 0; i < 4; ++i) {
                int newx = startx + d[i][0];
                int newy = starty + d[i][1];
                if (inArea(newx, newy) && !visited[newx][newy] && searchWord(board, word, index + 1, newx, newy))
                        return true;
            }
            visited[startx][starty] = false;
        }

        return false;
    }
public:
    bool exist(vector<vector<char>>& board, string word) {

        m = board.size();
        assert(m > 0);
        n = board[0].size();

        visited = vector<vector<bool>>(m, vector<bool>(n, false));

        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if (searchWord(board, word, 0, i, j))
                    return true;
            }
        }
        return false;
    }
};