LeetCode – Next Permutation (Python)

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.

The solution is the Pandit’s algorithm

1 Pandit’s algorithm
The article in Wikipedia describes the algorithm invented by Narayana Pandit to changes the list in-place to generate the next permutation given a list or Array A.

1. Find the largest index i such that A[i] < A[i + 1]. If no such index exists, the permutation is the last permutation. In this case, we just reverse the list to get the next permutation.
   
   Python

   #the largest index where A[i] < A[i + 1] # -1 if there is no such i where A[i] < A[i + 1] def get_split_index(A): i = len(A) - 1 while i > 0: if A[i - 1] >= A[i]: i -= 1 else: break return i - 1

   1 2 3 4 5 6 7 8 9 10

   #the largest index where A[i] < A[i + 1] # -1 if there is no such i where A[i] < A[i + 1] def get_split_index(A):     i = len(A) - 1     while i > 0:         if A[i - 1] >= A[i]:             i -= 1         else:             break     return i - 1

2. Find the largest index j greater than i such that A[j] > A[i].
   
   Python

   def get_chang_index(A, i): j = len(A) - 1 while j >= i: if A[j] > A[i]: break else: j -= 1 return j

   1 2 3 4 5 6 7 8

   def get_chang_index(A, i):     j = len(A) - 1     while j >= i:         if A[j] > A[i]:             break         else:             j -= 1     return j

3. Swap the value of A[i] with that of A[j].
   
   Python

   def swap(A, i, j): A[i], A[j] = (A[j], A[i])

   1 2 3

   def swap(A, i, j):     A[i], A[j] = (A[j], A[i])

4. Reverse the remaining sequence from A[i + 1] up to the end of the list.
   
   Python

   #reverse A[start:] def reverse(A, start): left = start right = len(A) - 1 while left < right: swap(A, left, right) left += 1 right -= 1

   1 2 3 4 5 6 7 8

   #reverse A[start:] def reverse(A, start):     left = start     right = len(A) - 1     while left < right:         swap(A, left, right)         left += 1         right -= 1

2 Example

Let’s put them together, and use A = [1, 2, 3, 4] to demonstrate the sequence of calling the get next permutation method.

The output is :