## 54. Spiral Matrix && 59. Spiral Matrix II

Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

```Input: 3
Output:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]```
```class Solution {
public:
vector<vector<int>> generateMatrix(int n) {

vector<vector<int>> res(n,vector<int>(n,0));
if (n == 0) return res;
int i = 1;
int rowS = 0,rowE = n - 1,colS = 0,colE = n -1;
while(i <= n * n)
{
for (int j = colS;j <= colE;++j)
{
res[rowS][j] = i++;
}
rowS++;

for (int j = rowS;j <= rowE;++j)
{
res[j][colE] = i++;
}
colE--;

for (int j = colE;j >= colS;--j)
{
res[rowE][j] = i++;
}
rowE--;

for (int j = rowE;j >= rowS;--j)
{
res[j][colS] = i++;
}
colS++;
}
return res;
}
};```

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]```
```class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
int m = matrix.size(), n = m ? matrix[0].size() : 0, u = 0, d = m - 1, l = 0, r = n - 1, p = 0;
vector<int> order(m * n);
while (u <= d && l <= r) {
for (int col = l; col <= r; col++) {
order[p++] = matrix[u][col];
}
if (++u > d) {
break;
}
for (int row = u; row <= d; row++) {
order[p++] = matrix[row][r];
}
if (--r < l) {
break;
}
for (int col = r; col >= l; col--) {
order[p++] = matrix[d][col];
}
if (--d < u) {
break;
}
for (int row = d; row >= u; row--) {
order[p++] = matrix[row][l];
}
if (l++ > r) {
break;
}
}
return order;
}
};```