Problem Statement:

Given an m x n matrix, return all elements of the matrix in spiral order.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

Solution:

Just remember to keep track of start and end indices for x and y directions.

 
class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) 
    {
        int m=matrix.size(), n=matrix[0].size(), i=0, j=0;
        int sy=0, sx=0, ey=m, ex=n; // m=ey=3, n=ex=4
        vector<int> res;
        while (res.size() < m*n)
        {
            while (res.size() < m*n && j<ex)
            {
                res.push_back(matrix[i][j]);
                j++;
            }
            j--; i++; ex--;
            while (res.size() < m*n && i<ey)
            {
                res.push_back(matrix[i][j]);
                i++;
            }
            i--; j--; ey--;
            while (res.size() < m*n && j>=sx)
            {
                res.push_back(matrix[i][j]);
                j--;
            }
            j++; i--; sx++;
            while (res.size() < m*n && i>sy)
            {
                res.push_back(matrix[i][j]);
                i--; 
            }
            i++; j++; sy++;
        }
        return res;
    }
};