# Spiral Matrix II   ## Intuition

This problem is not hard to get the algorithm. But its hard to notice all the edge cases.

So we need a good start. First, it's hard to combine all the 4 situation (left, right, up, down). So we would solve it by using 4 different if condition. And move one printer (at position matrix[x][y]) to mark the values one by one.

We would need 4 variables to mark the wall, lo_x, hi_x, lo_y and hi_y. Each time we hit the wall, we need to turn. And each time we turn, one line is finished, and we need to move a wall to make sure we don't hit it for more than once.

## Solution

``````
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
lo_x, lo_y, hi_x, hi_y = 0, 0, n - 1, n - 1
x, y = 0, 0
matrix = [[None] * n for _ in range(n)]
direction = 0
count = 1
while True:
if hi_x < lo_x or hi_y < lo_y:
return matrix

matrix[x][y] = count
count += 1
if direction == 0:
if y == hi_y:
lo_x += 1
direction = 1
x += 1
else:
y += 1
elif direction == 1:
if x == hi_x:
hi_y -= 1
direction = 2
y -= 1
else:
x += 1
elif direction == 2:
if y == lo_y:
hi_x -= 1
direction = 3
x -= 1
else:
y -= 1
else:
if x == lo_x:
lo_y += 1
direction = 0
y += 1
else:
x -= 1

``````