Leetcode: Spiral Matrix (0ms) Analysis and solution

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].

Analysis:

Logic is not very difficult, but it has lots of trivial things inside. The procedure can be divided into 4 sections and then process the four sections individually. The key here is the boundary control. See following figure for a clear logic and procedure to implement this problem:

In the figure, four index are used to mark the boundary: upper, right, bottom and left. The code according to above procedure:

//////////////////////////////////////////////////////////
// code 0ms

class Solution {

public:

       vector<int> spiralOrder(vector<vector<int>>& matrix) {

              int upper = 0, right = 0, bottom = 0, left = 0;

              vector<int> res;

              //check input

              if (matrix.size() == 0)return res;

              int m = matrix.size(), n = matrix[0].size(), index = 0, Size = m*n;

              while (index < Size + 1){

                     //section 1

                     for (int j = left; j < n - right; j++){

                           res.push_back(matrix[upper][j]);

                           ++index;

                     }

                     upper++;

                     if (index >= Size)return res;

                     //section 2

                     for (int i = upper; i < m - right; i++){

                           res.push_back(matrix[i][n - right - 1]);

                           ++index;

                     }

                     if (index >= Size)return res;

                     right++;

                     //section 3

                     for (int j = n - right - 1; j >= left; j--){

                           res.push_back(matrix[m-bottom-1][j]);

                           ++index;

                     }

                     if (index >= Size)return res;

                     bottom++;

                     //section 4

                     for (int i = m - bottom - 1; i > upper-1; i--){

                           res.push_back(matrix[i][left]);

                           ++index;

                     }

                     if (index >= Size)return res;

                     left++;

              }

              return res;

       }

};