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() returnRun Time: 56 ms
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