Given an m x n matrix, return all elements of the matrix in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[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]
Constraints:
Search
Searchm == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
ROWS, COLS = len(matrix), len(matrix[0])
top, left = 0, 0
right, bottom = COLS, ROWS
res = []
while left < right and top < bottom:
# get every i in top row
for i in range(left, right):
res.append(matrix[top][i])
top += 1
# get every i in right column
for i in range(top, bottom):
res.append(matrix[i][right - 1])
right -= 1
if not (top < bottom and left < right):
break
# get every i in bottom row
for i in range(right - 1, left - 1, -1):
res.append(matrix[bottom - 1][i])
bottom -= 1
# get every i in left col
for i in range(bottom - 1, top - 1, -1):
res.append(matrix[i][left])
left += 1
return res
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