dfs-矩阵题目

40. 顺时针打印矩阵

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。

class Solution {
public:
    vector<int> res;
    vector<vector<int> > mp;
    vector<vector<bool> > vis;
    int m, n;
    void dfs(int i, int j,int flag){
        if(i < 0 || i >= m || j < 0 || j >= n || vis[i][j]) return;
        vis[i][j] = 1;
        res.push_back(mp[i][j]);
        switch(flag){
            case 0: if(j < n-1 && !vis[i][j+1])
                    dfs(i,j+1,0);
                else dfs(i+1,j,1); break;
            case 1: if(i < m-1 && !vis[i+1][j])
                     dfs(i+1,j,1);
                else dfs(i,j-1,2); break;
            case 2:if(j >= 1 && !vis[i][j-1])
                dfs(i,j-1,2);
                else dfs(i-1,j,3);break;
            case 3:if(i >= 1 && !vis[i-1][j])
                dfs(i-1,j,3); else dfs(i,j+1,0);break;
        }
    }
    vector<int> printMatrix(vector<vector<int> > matrix) {
        mp = matrix;
        m = matrix.size();
        if(m == 0) return res;
        n = matrix[0].size();
        vis.resize(m);
        for(int i = 0; i < m; i++)
            vis[i].resize(n,0);
        dfs(0,0,0);
        return res;
    }
};

yxc的做法：定义四个方向以及改变方向的技巧：

class Solution {
public:
    vector<int> printMatrix(vector<vector<int>>& matrix) {
        vector<int> res;
        if (matrix.empty()) return res;
        int n = matrix.size(), m = matrix[0].size();
        vector<vector<bool>> st(n, vector<bool>(m, false));
        int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
        int x = 0, y = 0, d = 1;
        for (int k = 0; k < n * m; k ++ )
        {
            res.push_back(matrix[x][y]);
            st[x][y] = true;

            int a = x + dx[d], b = y + dy[d];
            if (a < 0 || a >= n || b < 0 || b >= m || st[a][b])
            {
                d = (d + 1) % 4;
                a = x + dx[d], b = y + dy[d];
            }
            x = a, y = b;
        }
        return res;
    }
};

作者：yxc
链接：https://www.acwing.com/solution/AcWing/content/748/
来源：AcWing
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