Next lexicographical permutation algorithm

Related problem

31. Next Permutation

A good explanation

Image is from: https://www.nayuki.io/page/next-lexicographical-permutation-algorithm

My code based on this algorithm

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        if len(nums)<=1:return nums
        j = len(nums)-2
        while j>=0:
            if nums[j]<nums[j+1]:
                break
            j -= 1
        else:
            nums.reverse()
            return
        pivot = j
        val = nums[j]
        this_min = float('inf')
        target_k = 0
        for k in range(j+1, len(nums)):
            if val<nums[k]<=this_min:
                target_k = k
        nums[pivot], nums[target_k] = nums[target_k], nums[pivot]
        l, r = pivot+1, len(nums)-1
        while l<r:
            nums[l], nums[r] = nums[r], nums[l]
            l+=1
            r-=1
        return

Code from a post

def nextPermutation(self, nums):
    i = j = len(nums)-1
    while i > 0 and nums[i-1] >= nums[i]:
        i -= 1
    if i == 0:   # nums are in descending order
        nums.reverse()
        return 
    k = i - 1    # find the last "ascending" position
    while nums[j] <= nums[k]:
        j -= 1
    nums[k], nums[j] = nums[j], nums[k]  
    l, r = k+1, len(nums)-1  # reverse the second part
    while l < r:
        nums[l], nums[r] = nums[r], nums[l]
        l +=1 ; r -= 1