class LinkedList:
    def __init__(self, value):
        self.value = value
        self.next = None


# return the head of the new linked list
def reverseLinkedList(head):
		# go!

We start out with a linked list like this

null   0 -> 1 -> 2 -> 3 -> 4 -> 5
prev   A    B

The variable head points at 0.

There are a few constraints when solving this problem

• Your solution should absolutely be O(n)
• Use as few variables as possible - management of the pointers is what makes this problem challenging

Walking through each iteration helped me.

// after 1 iteration
null <- 0    1 -> 2 -> 3 -> 4 -> 5
      prev   A    B

We set A.next to prev, then iterate forward. The variable B really just exists so we can iterate forward once we point A.next elsewhere.

// after 2 iterations
null <- 0 <- 1    2 -> 3 -> 4 -> 5
            prev  A    B 

// after 3 iterations
null <- 0 <- 1 <- 2    3 -> 4 -> 5
                 prev  A    B 

// after 4 iterations
null <- 0 <- 1 <- 2 <- 3    4 -> 5
                      prev  A    B 

// after 5 iterations
null <- 0 <- 1 <- 2 <- 3 <- 4    5
                           prev  A    B 

// after 6 iterations
null <- 0 <- 1 <- 2 <- 3 <- 4 <- 5    null null
                                 prev  A    B

Note that prev now points at the new head.

Here’s the final solution.

class LinkedList:
    def __init__(self, value):
        self.value = value
        self.next = None


def reverseLinkedList(head):
		prev = None
		a = head
		b = a.next
		
		while a is not None:
				a.next = prev
				
				prev = a
				a = b
				b = b.next if b is not None else None
			
		return prev

You can get rid of the if b is not None else None section if you rearrange the order in which you set everything (you set b = a.next before you do a.next = prev, but honestly I find that harder to reason about.

This solution has O(n) time complexity and O(1) space complexity.