dfs 返回值用bool相对void会快一点

力扣 剑指 Offer 12. 矩阵中的路径

超时代码
dfs返回值是void，用类内的全局变量flag表示找到或没找到。

class Solution {
public:
    bool flag;
    int vis[210][210];
    int dir[4][2] = { {1,0},{0,1},{-1,0},{0,-1} };
    int m, n;
    void dfs(int x, int y, int count, string& word, vector<vector<char>>& board) {
        if (board[x][y] != word[count]) {
            return;
        }
        if (count == word.size() - 1) {
            flag = true;
            return;
        }
        for (int i = 0; i < 4; i++) {
            int nx = x + dir[i][0];
            int ny = y + dir[i][1];
            if (nx >= 0 && nx <= m - 1 && ny >= 0 && ny <= n - 1 && vis[nx][ny] == 0) {
                vis[nx][ny] = 1;
                dfs(nx, ny, count + 1, word, board);
                vis[nx][ny] = 0;
            }
        }
    }

    bool exist(vector<vector<char>>& board, string word) {
        m = board.size();
        n = board[0].size();
        int sn = word.size();
        flag = false;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                memset(vis, 0, sizeof(vis));
                vis[i][j] = 1;
                dfs(i, j, 0, word, board);
                vis[i][j] = 0;
                if (flag) break;
            }
        }
        return flag;
    }
};

通过代码
将dfs返回值改为bool类型就通过了。分析原因可能是找到一条满足的路径后立即返回，如果没有返回值，还要将整个dfs执行完才能结束，可能递归调用和栈的一些操作浪费了一些时间（这从时间复杂度角度不好分析）

class Solution {
public:
    bool flag;
    int vis[210][210];
    int dir[4][2] = { {1,0},{0,1},{-1,0},{0,-1} };
    int m, n;
    bool dfs(int x, int y, int count, string& word, vector<vector<char>>& board) {
        if (board[x][y] != word[count]) {
            return false;
        }
        if (count == word.size() - 1) {
            flag = true;
            return true;
        }
        for (int i = 0; i < 4; i++) {
            int nx = x + dir[i][0];
            int ny = y + dir[i][1];
            if (nx >= 0 && nx <= m - 1 && ny >= 0 && ny <= n - 1 && vis[nx][ny] == 0) {
                vis[nx][ny] = 1;
                if (dfs(nx, ny, count + 1, word, board)) return true;
                vis[nx][ny] = 0;
            }
        }
        return false;
    }

    bool exist(vector<vector<char>>& board, string word) {
        m = board.size();
        n = board[0].size();
        flag = false;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                memset(vis, 0, sizeof(vis));
                vis[i][j] = 1;
                if (dfs(i, j, 0, word, board)) return true;
                vis[i][j] = 0;
                if (flag) return true;
            }
        }
        return false;
    }
};