算法描述：

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

Example 1:

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

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]
 

解題思路：模擬方法解決，註意邊界及細節。

    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> result;
        if(matrix.size()==0 || matrix[0].size() == 0)
            return result;
        int rowBegin = 0;
        int rowEnd = matrix.size()-1;
        int colBegin = 0
 
;
        int colEnd = matrix[0].size()-1;
        while(rowBegin <= rowEnd && colBegin <= colEnd){
            for(int i=colBegin; i<= colEnd; i++) result.push_back(matrix[rowBegin][i]);
            rowBegin++;
            for(int i=rowBegin; i<= rowEnd; i++) result.push_back(matrix[i][colEnd]);
            colEnd
 
--;
            if(colBegin <= colEnd && rowBegin <= rowEnd){
                for(int i = colEnd; i >= colBegin; i--) result.push_back(matrix[rowEnd][i]);
                rowEnd--;
                for(int i = rowEnd; i >= rowBegin; i--) result.push_back(matrix[i][colBegin]);
                colBegin++;
            }
        }
        return result;
    }

LeetCode-54-Spiral Matrix