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]

code

 

#include <iostream>
#include <vector>
using namespace std;

class Solution
{
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix)
    {
        vector<int> res;
        if(matrix.empty())
            return res;
        else if(matrix.size()==1)
            return matrix.at(0);

        int direction=0;
        int top=0,bottom=matrix.size()-1;
        int left=0,right=matrix.at(0).size()-1;
        while(true)
        {
            if(top>bottom||left>right)
                break;

            if(direction%4==0)
            {
                for(int i=left;i<=right;++i)
                    res.push_back(matrix.at(top).at(i));
                ++top;
            }
            else if(direction%4==1)
            {
                for(int i=top;i<=bottom;++i)
                    res.push_back(matrix.at(i).at(right));
                --right;
            }
            else if(direction%4==2)
            {
                for(int i=right;i>=left;--i)
                    res.push_back(matrix.at(bottom).at(i));
                --bottom;
            }
            else if(direction%4==3)
            {
                for(int i=bottom;i>=top;--i)
                    res.push_back(matrix.at(i).at(left));
                ++left;
            }
            ++direction;
        }
        return res;
    }
};

int main()
{
    vector<vector<int>> arr{{1,2,3},{4,5,6},{7,8,9}};
    Solution s;
    vector<int> res(s.spiralOrder(arr));
    for(auto i:res)
        cout<<i<<" ";
    cout<<endl;
    return 0;
}

 

 

vector<int> spiralOrder(vector<vector<int>>& matrix)
{
    if ( !matrix.size() || !matrix[0].size() )
        return vector<int>();

    vector<int> spiral(matrix.size()*matrix[0].size());

    vector<vector<int>> dirs = {{0,1},{1,0},{0,-1},{-1,0}};

    vector<int> numColsRows = { matrix[0].size(), matrix.size()-1 };

    vector<int> cellIndex = {0,-1};

    int nPasses = 0, k = 0;

    while ( k < spiral.size() )
    {
        // iterate to end of current column (%=0) or row (%=1)
        for (int i = 0; i < numColsRows[nPasses%2]; i++  )
        {
            // advance one cell and copy
            cellIndex[0] += dirs[nPasses%4][0];
            cellIndex[1] += dirs[nPasses%4][1];
            spiral[k++] = matrix[cellIndex[0]][cellIndex[1]];
        }

        // decrement number of rows or columns for next pass
        numColsRows[nPasses%2]--;
        nPasses++;
    }
    return spiral;
}