题意：给定一个m*n的矩阵，求在矩阵中是否能搜索到这个单词

https://leetcode.com/problems/word-search/description/
Input: board = [[“A”,“B”,“C”,“E”],[“S”,“F”,“C”,“S”],[“A”,“D”,“E”,“E”]], word = “ABCCED”
Output: true
解答：遇到第一个匹配的开头（比如例子中的A），开始搜索，用dfs来搜索四个方向，并用vis数组确保已经访问过的节点不会被访问，用一个变量来记录我当前搜索到哪一个位置了

class Solution {
public:
    bool ret;
    bool exist(vector<vector<char>>& board, string word) {
        if(!word.size()) {
            return true;
        }
        int m = board.size();
        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]) {
                    vector<vector<int>> vis(m, vector<int>(n,0));
                    vis[i][j] = 1;
                    dfs(board, 1, i, j, vis, word);
                    if(ret) {
                        return true;
                        }
                    }
                }
            }
            return false;
        }

    void dfs(vector<vector<char>>& board, int st, int x, int y, vector<vector<int>>& vis, string word) {
        int m = board.size();
        int n = board[0].size();
        if(st == word.size()) {
            ret = true;
            return;
        }
        
        int dk[] = {-1, 0, 1, 0, -1};
        for(int i = 0; i < 4; i++) {
            int dx = x + dk[i];
            int dy = y + dk[i+1];
            if(dx >= 0 && dx < m && dy >= 0 && dy < n && !vis[dx][dy] && board[dx][dy] == word[st]) {
                vis[dx][dy] = 1;
                dfs(board, st+1, dx, dy, vis, word);
                vis[dx][dy] = 0;
            }
        }
    }
};

question：为什么下面的代码过不了？以及有哪些不好的地方？

class Solution {
public:
    bool ret;
    bool exist(vector<vector<char>>& board, string word) {
        if(!word.size()) {
            return true;
        }
        int m = board.size();
        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]) {
                    vector<vector<int>> vis(m, vector<int>(n,0));
                    vis[i][j] = 1;
                    dfs(board, 1, i, j, vis, word);
                    if(ret) {
                        return true;
                        }
                    }
                }
            }
            return false;
        }

    void dfs(vector<vector<char>>& board, int st, int x, int y, vector<vector<int>> vis, string word) {
        int m = board.size();
        int n = board[0].size();
        if(st == word.size()) {
            ret = true;
            return;
        }
        
        int dk[] = {-1, 0, 1, 0, -1};
        for(int i = 0; i < 4; i++) {
            int dx = x + dk[i];
            int dy = y + dk[i+1];
            if(dx >= 0 && dx < m && dy >= 0 && dy < n && !vis[dx][dy] && board[dx][dy] == word[st]) {
                vis[dx][dy] = 1;
                dfs(board, st+1, dx, dy, vis, word);
                vis[dx][dy] = 0;
            }
        }
    }
};

为什么过不了？因为vis数组产生了copy，这个开销是MN的，你的答案的算法复杂度会变成M2N2
并且这个ret完全没必要，你完全可以通过dfs的返回值消掉这个ret
而且vis数组其实可以写在里面的

question：为什么这个会有问题？因为board[dx][dy] == word[st+1] 会有越界问题的，要想输入dfs(board, 0, i, j, vis, word,我必须每次进入的时候判定，而不是说每一次for循环中去判定

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(!word.size()) {
            return true;
        }
        int m = board.size();
        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]) {
                    vector<vector<int>> vis(m, vector<int>(n,0));
                    if(dfs(board, 0, i, j, vis, word)) {
                        return true;
                        }
                    }
                    
                }
            }
            return false;
        }

    bool dfs(vector<vector<char>>& board, int st, int x, int y, vector<vector<int>>& vis, string word) {
        int m = board.size();
        int n = board[0].size();
        vis[x][y] = 1;
        if(st == word.size()) {
            return true;
        }
        int dk[] = {-1, 0, 1, 0, -1};
        for(int i = 0; i < 4; i++) {
            int dx = x + dk[i];
            int dy = y + dk[i+1];
            if(dx >= 0 && dx < m && dy >= 0 && dy < n && !vis[dx][dy] && board[dx][dy] == word[st+1]) {
                if(dfs(board, st+1, dx, dy, vis, word)) {
                    return true;
                }
            }
        }
        vis[x][y] = 0;
        return false;
    }


};

最为简洁的写法,写的时候注意，先判断true，再来false

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(!word.size()) {
            return true;
        }
        int m = board.size();
        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]) {
                    vector<vector<int>> vis(m, vector<int>(n,0));
                    if(dfs(board, 0, i, j, vis, word)) {
                        return true;
                        }
                    }
                    
                }
            }
            return false;
        }

    bool dfs(vector<vector<char>>& board, int st, int x, int y, vector<vector<int>>& vis, string word) {
        int m = board.size();
        int n = board[0].size();
        if(st == word.size()) {
            return true;
        }
        if(x < 0 || x >= m || y < 0 || y >= n) {
            return false;
        }
        if(vis[x][y] || board[x][y] != word[st]) {
            return false;
        }
        vis[x][y] = 1;
        int dk[] = {-1, 0, 1, 0, -1};
        for(int i = 0; i < 4; i++) {
            int dx = x + dk[i];
            int dy = y + dk[i+1];
                if(dfs(board, st+1, dx, dy, vis, word)) {
                    return true;
                }
            }
        vis[x][y] = 0;
        return false;
        }

};

算法复杂度（MN3^L）
MN是因为第一个大循环，第二个dfs相当于搜索长度为L的字符串，有3个方向可以搜索（去除了回退哪一个）