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C++ Solution to Coding Challenge 59. Spiral Matrix II

Nhut Nguyen
Â·Oct 6, 2022Â·

Problem statement

Given a positive integer `n`, generate an `n x n` matrix filled with elements from `1` to `n^2` in spiral order.

Example 1

``````Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]
``````

Example 2

``````Input: n = 1
Output: [[1]]
``````

Constraints

• `1 <= n <= 20`.

Solution

1. Starting from the top left of the matrix.

2. Going along the spiral direction.

3. Put the value to the matrix, starting from `1`.

Code

``````#include <vector>
#include <iostream>
using namespace std;
enum Direction {RIGHT, DOWN, LEFT, UP};
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> m(n, vector<int>(n));
int bottom = n - 1;
int right = n - 1;
int top = 0;
int left = 0;
int row = 0;
int col = 0;
Direction d = RIGHT;
int a = 1;
while (top <= bottom && left <= right) {
m[row][col] = a++;
switch (d) {
case RIGHT:
if (col == right) {
top++;
d = DOWN;
row++;
} else {
col++;
}
break;
case DOWN:
if (row == bottom) {
right--;
d = LEFT;
col--;
} else {
row++;
}
break;
case LEFT:
if (col == left) {
bottom--;
d = UP;
row--;
} else {
col--;
}
break;
case UP:
if (row == top) {
left++;
d = RIGHT;
col++;
} else {
row--;
}
break;
}
}
return m;
}
void printResult(vector<vector<int>>& m) {
cout << "[";
for (auto& r : m) {
cout << "[";
for (int a : r) {
cout << a << ",";
}
cout << "]";
}
cout << "]\n";
}
int main() {
auto m = generateMatrix(3);
printResult(m);
m = generateMatrix(1);
printResult(m);
}
``````
``````Output:
[[1,2,3,][8,9,4,][7,6,5,]]
[[1,]]
``````

Complexity

• Runtime: `O(n^2)`, where `n x n` is the size of the matrix.

• Extra space: `O(1)`.

References

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