【剑指offer】Q15:链表中的倒数第K个结点

def FindKthToTail(head, k):
	if head == None: return None
	front = head; back = head
	steps = 0
	# get | front - head | = k
	while front and steps < k: 
		front = front.next
		steps += 1
	# less than k nodes 
	if front == None: return None

	while front:
		front = front.next
		back = back.next
	return back

与面试题37：两个链表的第一个公共结点使用的是同样的策略。