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59. 螺旋矩阵 II
给定一个正整数 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正方形矩阵。
示例:
输入: 3
输出:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
回顾上一题
O(N^2)
之前的解法是，先把第一个填进结果。这次，我们直接遍历，但是把最后一个要记得放入数组
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if len(matrix) == 0 or len(matrix[0]) == 0:
return []
n = len(matrix)
m = len(matrix[0])
x_directions = [0, 1, 0, -1]
y_directions = [1, 0, -1, 0]
bounds = [0, m - 1, n - 1, 0]
shrink_bounds = [1, -1, -1, 1]
res = []
x = y = 0
while bounds[0] <= bounds[2] and bounds[3] <= bounds[1]:
for k in range(4):
while bounds[0] <= x + x_directions[k] <= bounds[2] and bounds[3] <= y + y_directions[k] <= bounds[1]:
res.append(matrix[x][y])
x += x_directions[k]
y += y_directions[k]
bounds[k] += shrink_bounds[k]
res.append(matrix[x][y])
return res
模拟法
O(N^2)
class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
res = [[0 for j in range(n)] for i in range(n)]
x_directions = [0, 1, 0, -1]
y_directions = [1, 0, -1, 0]
bounds = [0, n - 1, n - 1, 0]
shrink_bounds = [1, -1, -1, 1]
x = y = 0
count = 1
while bounds[0] <= bounds[2] and bounds[3] <= bounds[1]:
for k in range(4):
while bounds[0] <= x + x_directions[k] <= bounds[2] and bounds[3] <= y + y_directions[k] <= bounds[1]:
res[x][y] = count
x += x_directions[k]
y += y_directions[k]
count += 1
bounds[k] += shrink_bounds[k]
res[x][y] = count
return res
–End–
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/spiral-matrix-ii/
Link：https://leetcode.com/problems/spiral-matrix-ii/