"""
48. Rotate Image
Medium
1115
100


You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]
Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]
 
"""

可以寫成O(1)的空間複雜度，當然時間複雜度還是O(N^2)
首先回顧一下高中的數學知識,二維座標系上的一點(m,n)順時針旋轉90度後坐標為(n,-m)
矩陣左上角座標(0,0),右下角座標(n-1, n-1),x軸向下，y軸向右,和我們平時看到的座標系只是旋轉了一下，
所以矩陣順時針旋轉和我們平時遇到的座標系順時針旋轉是相同的,高中時的知識仍然適用，
所以矩陣順時針是圍繞點((n-1)/2,(n-1)/2)旋轉的,
設矩陣內一點座標為(a,b),相對於軸心得座標是(a-(n-1)/2, b-(n-1)/2),
旋轉90度以後相對座標是(b-(n-1)/2, (n-1)/2-a)
所以矩陣內的絕對座標是(b,n-1-a)
再看矩陣裡的點q0,旋轉90度到了q1,再旋轉90度到了q2,再旋轉90度到了q3，再旋轉一次就會到達原點，
這四個相對應的點必須一起實現,必須將矩陣這樣均勻地分為4份，每四個對應的點分別在這四個區域內，方法很多,
我的方法是對於第i行:0<=i<n-1,取(i,j):i<=j<n-1-i
q0:(i,j)
q1:(j,n-1-i)
q2:(n-1-i,n-1-j)
q3:(n-1-j,i)

class Solution:
    def rotate(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        n = len(matrix)
        for i in range(0,(n-1)//2+1):
            for j in range(i,n-1-i):
                tmp 
 
= matrix[i][j]
                matrix[i][j], matrix[j][n-1-i], matrix[n-1-i][n-1-j], matrix[n-1-j][i] = matrix[n-1-j][i], matrix[i][j], matrix[j][n-1-i], matrix[n-1-i][n-1-j]

網上還有一種思路就是先上下翻轉再沿主對角線翻轉

(x,y)->(-x,y)->(y,-x)

class Solution:
    def rotate(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        matrix.reverse()
        n = len(matrix)
        for i in range(0,n):
            for j in range(i):
                matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]