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73 changes: 73 additions & 0 deletions C++/spiral_matrix.cpp
Original file line number Diff line number Diff line change
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/Problem: Given an m x n matrix, return all elements of the matrix in spiral order.
/*
Input:
n=3 m=3
Matriz:
2 3 5
19 23 7
17 13 11
Output:
vector => 2 3 5 7 11 13 17 19 23
*/
#include <iostream>
#include<vector>
#include<cmath>
using namespace std;
vector<int> SpiralOrder(vector<vector<int>> Mat,int n,int m) {
int val = 0,a[n][m];
int k = 0, l = 0;
/* k - starting row index
n - ending row index
l - starting column index
m - ending column index
i - iterator
*/
vector<int> v;
//Spiral Form Order: (right->down->left->up->right->..->end)
while (k < n && l < m){
for (int i = l; i < m; ++i){//Right
v.push_back( Mat[k][i]);
}
k++;
for (int i = k; i < n; ++i){//Down
v.push_back(Mat[i][m-1]);
}
m--;
if (k < n)
{
for (int i = m-1; i >= l; --i){//left
v.push_back(Mat[n-1][i]);
}
n--;
}
if (l < m)
{
for (int i = n-1; i >= k; --i){//up
v.push_back( Mat[i][l]);
}
l++;
}
}
return v;
}
int main()
{
int m,n,i,j=0;
cin>>n>>m;
vector<vector<int>> Mat;
for(i=0;i<n;i++){
vector<int>a;
for(j=0;j<m;j++){
int v;
cin>>v;
a.push_back(v);
}
Mat.push_back(a);
}
vector<int> ans=SpiralOrder(Mat,n,m);
for(auto x:ans){
cout<<x<<" ";
}cout<<endl;
return 0;
}
//Time Complexity: O(M*N). To traverse the matrix O(m*n) time is required.