1. 054 Spiral Matrix
Given an m x n matrix, return all elements of the matrix in spiral order.
1.1. Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
1.2. 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]
1.3. Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
2 Process
2.1 analysis
- Scan matrix first row, append to temp = []
- Remove maxtrix first row
- Rotate the new matrix by 90 degrees (anticlockwise) by 【LC 048】
- Repeat above steps
2.2 Code for Rotate Matrix
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything,
modify matrix in-place instead.
"""
temp = zip(*matrix[::-1])
matrix[:] = list(temp)
same result
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything,
modify matrix in-place instead.
"""
matrix_new = []
for i in range(len(matrix)):
temp = []
for j in range(-1, -len(matrix)-1, -1):
temp.append(matrix[j][i])
matrix_new.append(temp)
matrix[:] = matrix_new
Matrix:
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
result:
[1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]
2.3 Code for Spiral Matrix
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
nums = []
mrow = len(matrix)
mcol = len(matrix[0])
while mrow:
for i in range(mcol):
nums.append(matrix[0][i])
if mrow == 1:
break
matrix[:] = matrix[1:]
matrix[:] = list(zip(*matrix))[::-1]
mrow = len(matrix)
mcol = len(matrix[0])
return nums
Matrix:
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
result:
[1, 2, 5, 9, 6, 3, 4, 7, 10, 11, 8, 12]