Word Search leetCode,,

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[  ["ABCE"],  ["SFCS"],  ["ADEE"]]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

题目的意思是根据一个给定的字符矩阵，查找是否存在某个字符串，查找移动的规则是：只能左、右、上、下移动。

看到题目，首先得反应是回溯法，明显的回溯特征。明白了回溯法之后，应该就相对比较简单了。这里需要注意的是矩阵的每个字符只能访问一次，一次需要对访问过的字符做标记，自然就可以想到开一个标记矩阵来标记字符是否访问过，但是这样会发现无论是在时间还是在空间开销都很大，为了避免这种开销，那可以在原矩阵的基础上标记。

class Solution {public:    //回溯法    bool FindExist(vector<vector<char> > &board, string word, int wIndex, int i, int j, int m, int n)   {char tmpchar = word[wIndex];if ((board[i][j] == tmpchar) && (word.size() == wIndex + 1)){return true;}if ((board[i][j] == tmpchar) && (word.size() > wIndex + 1)){board[i][j] = '!';if ((i >= 1 && FindExist(board, word, wIndex + 1, i - 1, j, m, n)) ||    (j >= 1 && FindExist(board, word, wIndex + 1, i, j - 1, m, n)) ||    (i < m - 1 && FindExist(board, word, wIndex + 1, i + 1, j, m, n)) ||    (j < n - 1 && FindExist(board, word, wIndex + 1, i, j + 1, m, n))){return true;}board[i][j] = tmpchar; //回溯步骤}return false;   }        bool exist(vector<vector<char>>& board, string word) {        int m = board.size();        if(m == 0)        {            return false;        }                int n = board[0].size();        for(int i = 0; i < m; ++i)        {            for(int j = 0; j < n; ++j)            {                if(board[i][j] == word[0])                {                    bool tmpRes = FindExist(board, word, 0, i, j, m, n);                    if(tmpRes)                    {                        return true;                    }                }            }        }                return false;    }};