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The description of the problem

The simplified version：
Given a $n\times n$ matrix, output its elements in spiral and chronological order.

Given a $m\times n$ matrix, return all elements of the matrix in spiral order.

一、The codes 1 for problem_!

#include 
#include 
using namespace std;
class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;
        int start_x = 0, start_y = 0;
        int offset = 1;
        int n = matrix.size();
        int i, j;
        for (int k = 0; k < n / 2; k++) {
            for (j = start_y; j < n - offset; j++) {
                res.push_back(matrix[start_x][j]);
            }
            for (i = start_x; i < n - offset; i++) {
                res.push_back(matrix[i][j]);
            }
            for (; j > start_y; j--) {
                res.push_back(matrix[i][j]);
            }
            for (; i > start_x; i--) {
                res.push_back(matrix[i][j]);
            }
            start_x = start_x + 1;
            start_y = start_y + 1;
            offset++;
        }
        if (n % 2) {
            res.push_back(matrix[n/2][n/2]);
        }
        return res;
    }   
};
int main()
{
    Solution s;
    vector<vector<int>> matrix = {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
    vector<int> res = s.spiralOrder(matrix);
    for (int i = 0; i < res.size(); i++) {
        cout << res[i] << " ";
    }
    cout << endl;
    return 0;
}
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The codes for the second problem

#include 
#include 
using namespace std;
class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;
        int m = matrix.size();
        int n = matrix[0].size();
        int start_x = 0, start_y = 0;
        int offset = 1;
        int i = 0, j = 0;
        int layers = m < n ? m : n;
        layers = layers / 2;
        for (int k = 0; k < layers; k++) {
            for (j = start_y; j < n - offset; j++) {
                res.push_back(matrix[start_x][j]);
            }
            for (i = start_x; i < m - offset; i++) {
                res.push_back(matrix[i][j]);
            }
            for (; j > start_y; j--) {
                res.push_back(matrix[i][j]);
            }
            for (; i > start_x; i--) {
                res.push_back(matrix[i][j]);
            }
            start_x++;
            start_y++;
            offset++;
        }
        if ((m <= n) && (m % 2 == 1)) {
            for (int k = 0; k < (n - 2*layers); k++) {
                res.push_back(matrix[start_x][start_y + k]);
            }
        } else if ((n < m) && (n % 2 == 1)) {
            for (int k = 0; k < (m - 2*layers); k++) {
                res.push_back(matrix[start_x + k][start_y]);
            }
        }
        return res;
    }
};
int main() {
    vector<vector<int>> matrix = {
        {1, 2, 3, 4},
        {5, 6, 7, 8},
        {9, 10, 11, 12},
        {13, 14, 15, 16}
    };
    Solution s;
    vector<int> res = s.spiralOrder(matrix);
    for (int i = 0; i < res.size(); i++) {
        cout << res[i] << " ";
    }
    cout << endl;
    return 0;
}
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conclusion

know your variables and use them correctly.