让a是n×n的阵列基于0的索引

f = floor(n/2)
c = ceil(n/2)

for x = 0 to f - 1
  for y = 0 to c - 1
    temp = a[x,y]
    a[x,y] = a[y,n-1-x]
    a[y,n-1-x] = a[n-1-x,n-1-y]
    a[n-1-x,n-1-y] = a[n-1-y,x]
    a[n-1-y,x] = temp

编辑 如果您想避免使用temp，这一次可以在python中工作（它也会以正确的方向旋转）。

def rot2(a):
  n = len(a)
  c = (n+1) / 2
  f = n / 2
  for x in range(c):
    for y in range(f):
      a[x][y] = a[x][y] ^ a[n-1-y][x]
      a[n-1-y][x] = a[x][y] ^ a[n-1-y][x]
      a[x][y] = a[x][y] ^ a[n-1-y][x]

      a[n-1-y][x] = a[n-1-y][x] ^ a[n-1-x][n-1-y]
      a[n-1-x][n-1-y] = a[n-1-y][x] ^ a[n-1-x][n-1-y]
      a[n-1-y][x] = a[n-1-y][x] ^ a[n-1-x][n-1-y]

      a[n-1-x][n-1-y] = a[n-1-x][n-1-y]^a[y][n-1-x]
      a[y][n-1-x] = a[n-1-x][n-1-y]^a[y][n-1-x]
      a[n-1-x][n-1-y] = a[n-1-x][n-1-y]^a[y][n-1-x]

注意：这仅适用于整数矩阵。