leetcode原题链接：单词搜索

题目描述

       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

解题方法：回溯。用backtrack(board, x, y, word, i, visited)表示是否存在以board[x][y]为开始节点，能找到单词word，为了防止重复兜圈查找，用visited变量保存标记当前已经访问的位置。

核心逻辑如下：

1）visited[x][y]=true，比较board[x][y]是否等于word[i]，不相等直接返回false；

2)   如果board[x][y]等于word[i]，则继续向上、下、左、右四个方向继续搜索word[i+1]；

3）结果为: result(上) || result(下) || result(左) || result(右）

4）visited[x][y]=false 恢复状态

C++代码

#include <iostream>
#include <vector>
#include <string>

class Solution {
public:
    bool exist(std::vector<std::vector<char>>& board, std::string word) {
        int rnum = board.size();
        int cnum = board[0].size();
        std::vector<std::vector<bool>> visited(rnum, std::vector<bool>(cnum, false));
        
        for (int x = 0; x < rnum; x++) {
            for (int y = 0; y < cnum; y++) {
                if (backtrack(board, x, y, word, 0, visited)) { //以(x,y)为开始节点去搜索
                    return true;
                }
            }
        }
        return false;
    }

    //判断以board(x,y)为起点是否能找到word
    bool backtrack(std::vector<std::vector<char>>& board, int x, int y, 
                   std::string word, int i, std::vector<std::vector<bool>>& visited) {
        int rnum = board.size();
        int cnum = board[0].size();
        int word_n = word.size();
        if (i == word_n) {
            return true;
        }
        if (x < 0 || x >= rnum || y < 0 || y >= cnum) {
             return false;
        }
        if (visited[x][y]) {//已经访问过，直接返回
            return false;
        }
        if (board[x][y] != word[i]) { //首字母不同，直接返回
            return false;
        }
        visited[x][y] = true;//标记为访问
        bool result = false;
        result = backtrack(board, x - 1, y, word, i + 1, visited) || //向上查找
                 backtrack(board, x + 1, y, word, i + 1, visited) || //向上
                 backtrack(board, x, y - 1, word, i + 1, visited) || //向左
                 backtrack(board, x, y + 1, word, i + 1, visited);   //向右
        visited[x][y] = false;
        return result;
    }
};