Chapter 2 Linked Lists

Problem 2.2: Implement an algorithm to find the nth to last element of a singly linked list.

The solution on answer page needs only one iteration while other intuitive solution takes more than one iteration.
def find_nth_to_last(head, n):
    p1 = p2 = head
    # Move p2 n nodes forward
    for i in range(0, n):
        # If p2 runs into the end of list,
        # the input is invalid
        if p2 == None:
            return False
        p2 = p2.next
    # Move p2 and p1 together until
    # p2 runs into the end of list
    while True:
        if p2 == None:
            return p1
        p2 = p2.next
        p1 = p1.next
I learned that sometimes we can use more than one pointers to achieve a better solution.;