Spiral Matrix I, II

I. Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

You should return [1,2,3,6,9,8,7,4,5].

Analyse: four steps. Top right, right down, bottom left, left up.

Runtime: 0ms. 

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size();
        vector<int> result;
        if(m == 0) return result;
        
        int n = matrix[0].size();
        int x = 0, y = 0;
        while(m > 0 && n > 0){
            if(m == 1){
                for(int i = 0; i < n; i++)
                    result.push_back(matrix[x][y++]);
                break;
            }
            else if(n == 1){
                for(int j = 0; j < m; j++)
                    result.push_back(matrix[x++][y]);
                break;
            }
            
            for(int i = 0; i < n - 1; i++){//top-right scan
                result.push_back(matrix[x][y++]);
            }
            for(int i = 0; i < m - 1; i++){//right-down scan
                result.push_back(matrix[x++][y]);
            }
            for(int i = 0; i < n - 1; i++){//bottom-left scan
                result.push_back(matrix[x][y--]);
            }
            for(int i = 0; i < m - 1; i++){//left-up scan
                result.push_back(matrix[x--][y]);
            }
            
            x++;
            y++;
            m -= 2;
            n -= 2;
        }
        return result;
    }
};

II. Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

Analyse: The same as 1. 

Runtime: 8ms. 

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int> > result(n, vector<int>(n, 0));
        if(n == 0) return result;
        if(n == 1){
            result[0][0] = 1;
            return result;
        }
        
        int current = 0;
        int x = 0, y = 0;
        int row = n, col = n;
        while(row > 0 && col > 0){
            if(row == 1){
                for(int i = 0; i < col; i++)
                    result[x][y++] = ++current;
                break;
            }
            else if(col == 1){
                for(int i = 0; i < row; i++)
                    result[x++][y] = ++current;
                break;
            }
            
            for(int i = 0; i < col - 1; i++){//put top-right
                result[x][y++] = ++current;
            }
            for(int i = 0; i < row - 1; i++){//put top-down
                result[x++][y] = ++current;
            }
            for(int i = 0; i < col - 1; i++){//put bottom-left
                result[x][y--] = ++current;
            }
            for(int i = 0; i < row - 1; i++){//put bottom-up
                result[x--][y] = ++current;
            }
            x++;
            y++;
            row -= 2;
            col -= 2;
        }
        return result;
    }
};