1 问题
给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。
示例 1：
输入：n = 3
输出：[[1,2,3],[8,9,4],[7,6,5]]
示例 2：
输入：n = 1
输出：[[1]]
2 答案
自己写的，参考54.螺旋矩阵
python
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
matrix = [[0 for _ in range(n)] for _ in range(n)]
up, down, left, right = 0, n-1, 0, n-1
x, y, cur_d = 0, 0, 0
dire = [[0, 1], [1, 0], [0, -1], [-1, 0]]
for i in range(n**2):
matrix[x][y] = i+1
if cur_d == 0 and y == right:
cur_d += 1
up += 1
if cur_d == 1 and x == down:
cur_d += 1
right -= 1
if cur_d == 2 and y == left:
cur_d += 1
down -= 1
if cur_d == 3 and x == up:
cur_d += 1
left += 1
cur_d %= 4
x += dire[cur_d][0]
y += dire[cur_d][1]
return matrix
官方解，循环用的while，方法相似
python
class Solution(object):
def generateMatrix(self, n):
if n == 0: return []
res = [[0] * n for i in range(n)]
left, right, up, down = 0, n - 1, 0, n - 1
x, y = 0, 0
dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
cur_d = 0
count = 0
while count != n * n:
res[x][y] = count + 1
count += 1
if cur_d == 0 and y == right:
cur_d += 1
up += 1
elif cur_d == 1 and x == down:
cur_d += 1
right -= 1
elif cur_d == 2 and y == left:
cur_d += 1
down -= 1
elif cur_d == 3 and x == up:
cur_d += 1
left += 1
cur_d %= 4
x += dirs[cur_d][0]
y += dirs[cur_d][1]
return res也可以利用坐标是否超过边界来变换遍历方向
python
class Solution(object):
def generateMatrix(self, n):
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
res = [[0] * n for i in range(n)]
x, y = 0, 0
count = 0
cur_d = 0
while count != n * n:
res[x][y] = count + 1
count += 1
dx, dy = directions[cur_d][0], directions[cur_d][1]
newx, newy = x + dx, y + dy # newx, newy 用于试错，看坐标是否超过边界
if newx < 0 or newx >= n or newy < 0 or newy >= n or res[newx][newy] != 0:
cur_d = (cur_d + 1) % 4
dx, dy = directions[cur_d][0], directions[cur_d][1]
x, y = x + dx, y + dy
return res