Reverse segments of a singly linked list by manipulating pointers — O(1) extra memory.

Data Flow

before:  1 -> 2 -> 3 -> 4 -> 5 -> None

iteratively:
   save next, flip cur.next to prev, advance prev/cur

after:   5 -> 4 -> 3 -> 2 -> 1 -> None

When to use

• Reverse linked list in place.
• Reverse every K-group (Reverse Nodes in K-Group).

Code (reverse entire list)

def reverse(head):
    prev, cur = None, head
    while cur:
        nxt = cur.next
        cur.next = prev
        prev = cur
        cur = nxt
    return prev

Pitfalls

• Losing next before the swap — if you overwrite cur.next first, the rest of the list is lost.
• Forgetting to stitch the previous segment to the reversed head.

Analogy

A train reversing engine order by detaching cars in pairs and re-attaching them in the opposite direction, one at a time.

Interview tip: Always use three pointers: prev, cur, nxt. The moment you forget to set cur.next = prev BEFORE updating prev, the list loops back.