First perform a transposition, then reverse every row.
1 2 1 3 3 1
3 4 => 2 4 => 4 2
Solution:
# T:O(n^2) S:O(1)
class Solution:
# @param {integer[][]} matrix
# @return {void} Do not return anything, modify matrix in-place instead.
def rotate(self, matrix):
n = len(matrix)
for i in xrange(n):
for j in xrange(i+1, n):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
# get transposition
for i in xrange(n):
matrix[i].reverse()
return
Run Time: 56 ms
No comments:
Post a Comment