【剑指offer】面试题37：两个链表的第一个公共结点

#@ util function, get the number of nodes in list referenced by head
def getLenOfList(head):
	nodeNum = 0
	while head:
		nodeNum += 1
		head = head.next
	return nodeNum

# find the first common node of two lists referenced by head1 and head2
def FindFirstCommonNode(head1, head2):
	if None == head1 or None == head2:
		return None
	lenList1 = getLenOfList(head1)
	lenList2 = getLenOfList(head2)

	#@ we make head1 point to the longer list
	if lenList2 > lenList1:
		head1, head2 = head2, head1

	#@ head1 skip |lenList1 - lenList2| nodes
	for i in range(lenList1 - lenList2):
		head1 = head1.next
	while None != head1:
		if head1 == head2:
			return head1
		head1 = head1.next
		head2 = head2.next

类似的题，推断给定的两个链表是否存在公共的结点，也就是是否在某个结点处两个链表汇聚。

思路是。假设汇聚的话，那么最后一个结点肯定是同样的，由于是单向链表，汇聚后，就不可能再出现分叉。

# judge wether two lists has common node or not, or if they crossed in some node
def IfHasCommonNode(head1, head2):
	# head1 move to the last noNone node
	while head1 and head1.next:
		head1 = head1.next
	# head2 move to the last noNone node
	while head2 and head2.next:
		head2 = head2.next

	# if the last node is same
	if head1 and head1 == head2:
		return True

	return False

再一个类似的题，推断链表是否存在环，复杂一点的，若存在环，则输出出现环的第一个结点。有兴趣的能够练习下。