Description
如何判断两个有环单链表是否相交？相交的话返回第一个相交的节点，不想交的话返回空。如果两个链表长度分别为N和M，请做到时间复杂度O(N+M)，额外空间复杂度O(1)。
给定两个链表的头结点head1和head2(注意，另外两个参数adjust0和adjust1用于调整数据,与本题求解无关)。请返回一个bool值代表它们是否相交。
Solutions
1. Different situation
要分几个情况讨论！
# -*- coding:utf-8 -*-
# Time: O(n)
# Space: O(1)
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class ChkIntersection:
def detectCycle(self, head):
if not head:
return None
fast, slow = head, head
while fast and slow:
if fast.next:
fast = fast.next.next
else:
return None
slow = slow.next
if fast == slow:
break
if not fast or not slow:
return None
restart = head
while restart != slow:
restart = restart.next
slow = slow.next
return slow
def chkInter(self, head1, head2, adjust0, adjust1):
point1 = self.detectCycle(head1)
point2 = self.detectCycle(head2)
node1, node2 = head1, head2
n1, n2 = 0, 0
while node1 != point1:
node1 = node1.next
n1 += 1
while node2 != point2:
node2 = node2.next
n2 += 1
diff = abs(n1 - n2)
node1, node2 = head1, head2
if n1 > n2:
for _ in range(diff):
node1 = node1.next
else:
for _ in range(diff):
node2 = node2.next
while node1 != point1 and node2 != point2:
if node1 == node2:
return node1
node1 = node1.next
node2 = node2.next
node1 = point1
while True:
if node1 == point2:
if n1 > n2:
return point2
else:
return point1
node1 = node1.next
if node1 == point1:
break
return False