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]
解题思路：模拟方法解决，注意边界及细节。
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; }