easy https://leetcode.com/problems/intersection-of-two-linked-lists/
题目描述：
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
Notes:
- If the two linked lists have no intersection at all, return
null. - The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
- Your code should preferably run in O(n) time and use only O(1) memory.
方法一：分别遍历 A 和 B 链表得到长度，较长的那个先走长度差值的 node 数，然后 A、B 一起走，看是否有相遇
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def getIntersectionNode(self, headA, headB): 9 """ 10 :type head1, head1: ListNode 11 :rtype: ListNode 12 """ 13 if not headA or not headB: 14 return None 15 lenA, lenB = 0,0 16 temp = headA 17 while temp: 18 temp = temp.next 19 lenA += 1 20 temp = headB 21 while temp: 22 temp = temp.next 23 lenB += 1 24 if lenA>=lenB: 25 for i in range(lenA-lenB): 26 headA = headA.next 27 else: 28 for i in range(lenB-lenA): 29 headB = headB.next 30 while headA and headB: 31 if headA==headB: 32 return headA 33 headA = headA.next 34 headB = headB.next 35 return None
方法二：利用题目中的链表无环的属性，A、B 同时从头出发，走到末尾处就转入对方的头节点，这样相遇的时候必然是走了相同的长度。若在尾部相遇则没有交叉节点，此时 A 和 B 均走了 len(A)+len(b) 的长度。若在中间相遇则相遇位置为交叉节点，此时 A 和 B 均走了 uniqueA+uniqueB+mutualAB 的长度。
1 class Solution(object): 2 def getIntersectionNode(self, headA, headB): 3 """ 4 :type head1, head1: ListNode 5 :rtype: ListNode 6 """ 7 if not headA or not headB: 8 return None 9 a,b = headA,headB 10 while a!=b: 11 if a: 12 a = a.next 13 else: 14 a = headB 15 if b: 16 b = b.next 17 else: 18 b = headA 19 return a