quickperm全排列算法

def permutation(n): # return 2d list
    if n == 1:
        return [[1]]

    lst = []

    for perm in permutation(n-1):
        for i in range(n):
            lst.append(perm[:i]+[n]+perm[i:])
    return lst

The Counting QuickPerm Algorithm:

   let a[] represent an arbitrary list of objects to permute
   let N equal the length of a[]
   create an integer array p[] of size N to control the iteration       
   initialize p[0] to 0, p[1] to 0, p[2] to 0, ..., and p[N-1] to 0
   initialize index variable i to 1
   while (i < N) do {
      if (p[i] < i) then {
         if i is odd, then let j = p[i] otherwise let j = 0
         swap(a[j], a[i])
         increment p[i] by 1
         let i = 1 (reset i to 1)
      } // end if
      else { // (p[i] equals i)
         let p[i] = 0 (reset p[i] to 0)
         increment i by 1
      } // end else (p[i] equals i)
   } // end while (i < N)