LeetCode Spiral Matrix

Description

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].

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

/*
*Spiral Matrix 
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<vector>
using namespace std;

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int rowBegin, rowEnd;
        int colBegin, colEnd;
        vector<int> result;
        int rowSize = matrix.size();
        if(rowSize<=0)
            return result;
            
        int colSize = matrix[0].size();
        if(colSize<=0)
            return result;
        
        rowBegin = 0; rowEnd = rowSize-1;
        colBegin = 0; colEnd = colSize-1;
        while(rowBegin<rowEnd && colBegin<colEnd) {
            for(int i=colBegin;i<colEnd;i++)
                result.push_back(matrix[rowBegin][i]);
            for(int i=rowBegin;i<rowEnd;i++)
                result.push_back(matrix[i][colEnd]);
            for(int i=colEnd;i>colBegin;i--)
                result.push_back(matrix[rowEnd][i]);
            for(int i=rowEnd;i>rowBegin;i--)
                result.push_back(matrix[i][colBegin]);
            rowBegin ++; rowEnd --;
            colBegin ++; colEnd --;
        }
        if(rowBegin == rowEnd && colBegin == colEnd)
            result.push_back(matrix[rowBegin][colBegin]);
        else{
            if(rowBegin == rowEnd){
                for(int i=colBegin;i<=colEnd;i++)
                    result.push_back(matrix[rowBegin][i]);
            }
            if(colBegin == colEnd){
                for(int i=rowBegin;i<=rowEnd;i++)
                    result.push_back(matrix[i][colBegin]);
            }
        }
        return result;
    }
};

Note

To solve the problem, four anchors are needed: rowBegin, rowEnd, colBegin, colEnd. The order I traveral the matrix: rowBegin, colEnd, rowEnd, colBegin. In the end, only one row or one column is remained, then traversal it.