LeetCodeOJ——Word Search

题目：
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 =

[
[‘A’,’B’,’C’,’E’],
[‘S’,’F’,’C’,’S’],
[‘A’,’D’,’E’,’E’]
]
word = “ABCCED”, -> returns true,
word = “SEE”, -> returns true,
word = “ABCB”, -> returns false.

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代码：

class Solution {public:    bool exist(vector<vector<char>>& board, string word) {        bool flag = false;        int size = board.size();        char road_val;        string road, dir;        stack<char>  word_stk, tmp;        int word_len = word.length();        for (unsigned i = 0; i <word_len; i++)        {            word_stk.push(word[word_len - 1 - i]);        }        for (vector<vector<char>>::size_type i = 0; i < size; i++)        {            for (vector<char>::size_type j = 0; j < size; j++)            {                road = "";                tmp = word_stk;                dir = "";                if (board[i][j] == word[0])                {                    road_val = i*size+j+'0';                    road = road + road_val;                    tmp.pop();                    if (fun(board, tmp, road, road_val, dir, i*size + j, size)) //查找成功                    {                        flag = true;                        return flag;                    }                }            }        }        return flag;    }    bool  fun(vector<vector<char>> board, stack<char> &word_stk, string &road, char road_val, string dir, int pos, int size)    {        int row, col;        bool flag = false;        row = pos / size;        col = pos % size;        //char road_val;        if (word_stk.empty())        {            return true;        }        if ((row<0 || row > size) || (col< 0 || col > size))        {            return false;        }    //  if (road.find(road_val) != string::npos && ( !road.empty()))    //  {    //      return false;    //  }        //up方向        if (row - 1 > 0 && dir != "up")        {            road_val = (row - 1)*size + col + '0';            if (road.find(road_val) == string::npos )            {                if (board[row - 1][col] == word_stk.top())  //匹配成功                {                    dir = "bottom";                    road += road_val;                    word_stk.pop();                    flag = fun(board, word_stk, road, road_val, dir, pos - size, size);                    if (flag)                    {                        return flag;                    }                    word_stk.push(board[row - 1][col]);                    road.erase(road.size() - 1, 1);                }            }        }        //left方向        if (col - 1 > 0 && dir != "left")        {            road_val = row * size + col - 1 + '0';            if (road.find(road_val) == string::npos)            {                if (board[row][col - 1] == word_stk.top())  //匹配成功                {                    dir = "right";                    road += road_val;                    word_stk.pop();                    flag = fun(board, word_stk, road, road_val, dir, pos - 1, size);                    if (flag)                    {                        return flag;                    }                    word_stk.push(board[row][col - 1]);                    road.erase(road.size() - 1, 1);                }            }        }        //bottom方向        if (row + 1 < size && dir != "bottom")        {            road_val = (row + 1)*size + col + '0';            if (road.find(road_val) == string::npos)            {                if (board[row + 1][col] == word_stk.top())  //匹配成功                {                    dir = "up";                    road += road_val;                    word_stk.pop();                    flag = fun(board, word_stk, road, road_val, dir, pos + size, size);                    if (flag)                    {                        return flag;                    }                    word_stk.push(board[row + 1][col]);                    road.erase(road.size() - 1, 1);                }            }        }        //right方向        if (col + 1 < size && dir != "right")        {            road_val = row * size + col + 1 + '0';            if (road.find(road_val) == string::npos)            {                if (board[row][col + 1] == word_stk.top())  //匹配成功                {                    dir = "left";                    road += road_val;                    word_stk.pop();                    flag = fun(board, word_stk, road, road_val, dir, pos + 1, size);                    if (flag)                    {                        return flag;                    }                    word_stk.push(board[row][col + 1]);                    road.erase(road.size() - 1, 1);                }            }        }           return flag;    }};

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结果：

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思考：
泪目~，/(ㄒoㄒ)/~~。总是超时，暂时还找不到另外一种思路。。。决定过一段时间再回头看看，能不能想出其他的思路，哎。。。