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#@ util function, get the number of nodes in list referenced by headdef 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 head2def 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，判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，如果汇聚的话，那么最后一个结点肯定是相同的，因为是单向链表，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 nodedef 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

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