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난이도 - Medium
Intuition
Transpose하고 y-axis swap하면 쉽게 풀 수 있다.
Code
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
n = len(matrix)
# Transpose
for i in range(n):
for j in range(i, n):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
# Swap in y-axis
for i in range(n):
for j in range(n//2):
matrix[i][j], matrix[i][n-j-1] = matrix[i][n-j-1], matrix[i][j]
Complexity
Time Complexity: O(M) 여기서 M은 cell의 개수이다.
Space Complexity: O(1)
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