三个月算法进阶--day60
目录
旋转矩阵
leetcode剑指offer第29题顺时针打印矩阵
旋转法
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
ans = []
while matrix:
ans.extend(matrix[0])
matrix = list(zip(* matrix[1:]))[::-1]
return ans
模拟路径(辅助矩阵或四个边界值)
class Solution:
def spiralOrder(self, matrix:[[int]]) -> [int]:
if not matrix: return []
l, r, t, b, res = 0, len(matrix[0]) - 1, 0, len(matrix) - 1, []
while True:
for i in range(l, r + 1): res.append(matrix[t][i]) # left to right
t += 1
if t > b: break
for i in range(t, b + 1): res.append(matrix[i][r]) # top to bottom
r -= 1
if l > r: break
for i in range(r, l - 1, -1): res.append(matrix[b][i]) # right to left
b -= 1
if t > b: break
for i in range(b, t - 1, -1): res.append(matrix[i][l]) # bottom to top
l += 1
if l > r: break
return res
层层剥开(非递归)
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if not matrix or not matrix[0]:
return list()
rows, columns = len(matrix), len(matrix[0])
order = list()
left, right, top, bottom = 0, columns - 1, 0, rows - 1
while left <= right and top <= bottom:
for column in range(left, right + 1):
order.append(matrix[top][column])
for row in range(top + 1, bottom + 1):
order.append(matrix[row][right])
if left < right and top < bottom:
for column in range(right - 1, left, -1):
order.append(matrix[bottom][column])
for row in range(bottom, top, -1):
order.append(matrix[row][left])
left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1
return order
python二维矩阵初始化
visited = [[0] * col for _ in range(row)]