Problem description:

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

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Example 1:

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

Input:
[
  [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]

Solution:
This is a very simple and easy to understand solution. I traverse right and increment rowBegin, then traverse down and decrement colEnd, then I traverse left and decrement rowEnd, and finally I traverse up and increment colBegin.

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class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;
        if(matrix.size()== 0) return res;
        if(matrix[0].size()== 0) return res;
        
        int rowBegin = 0;
        int rowEnd = matrix.size()-1;
        int colBegin = 0;
        int colEnd = matrix[0].size() - 1;
        
        while (rowBegin <= rowEnd && colBegin <= colEnd) {
            // Traverse Right
            for (int j = colBegin; j <= colEnd; j ++) {
                res.push_back(matrix[rowBegin][j]);
            }
            rowBegin++;
            
            // Traverse Down
            for (int j = rowBegin; j <= rowEnd; j ++) {
                res.push_back(matrix[j][colEnd]);
            }
            colEnd--;
            
            if (rowBegin <= rowEnd) {
                // Traverse Left
                for (int j = colEnd; j >= colBegin; j --) {
                    res.push_back(matrix[rowEnd][j]);
                }
            }
            rowEnd--;
            
            if (colBegin <= colEnd) {
                // Traver Up
                for (int j = rowEnd; j >= rowBegin; j --) {
                    res.push_back(matrix[j][colBegin]);
                }
            }
            colBegin ++;
        }
        
        return res;      
    }
};