Project Euler#24 Lexicographic permutations

#A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, #2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. #The lexicographic permutations of 0, 1 and 2 are:

#012   021   102   120   201   210

#What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

#Analysis
#The following algorithm generates the next permutation lexicographically
#after a given permutation. It changes the given permutation in-place.

#1. Find the largest index k such that a[k] < a[k + 1]. If no such
#   index exists, the permutation is the last permutation.
#2. Find the largest index l such that a[k] < a[l]. Since k + 1
#   is such an index, l is well defined and satisfies k < l.
#3. Swap a[k] with a[l].
#4. Reverse the sequence from a[k + 1] up to and including the final
#   element a[n].

#Source: http://en.wikipedia.org/wiki/Permutation

def perLexi(a):
    largestK = 0
    for k in range(len(a)-1):
        if a[k] < a[k+1] and k >= largestK :
            largestK = k

    if largestK == len(a) - 1:
        return a

    largestL = 0
    for l in range(len(a)):
        if a[largestK] < a[l] and l >= largestL :
            largestL = l

    a[largestK], a[largestL] = a[largestL], a[largestK]
    temp = a[largestK+1:]
    temp.reverse()
    for i in a[largestK+1:]:
        a.remove(i)
    for j in temp:
        a.append(j)

    return a

s = [0,1,2,3,4,5,6,7,8,9]

for i in range(999999):
    s = perLexi(s)
print s