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:
A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3
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.
解题思路 :
主要是考虑两条链路不一样长的问题,那么在其中一条链路到底的时候立马换回到另一条链路上面,另一条链路到底的时候也换回到这一条链路上,这样保证了后面能够同时到达相交点。
代码如下:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { if(headA == null || headB == null) return null; ListNode a = headA; ListNode b = headB; while(a != b){ if(a == null) a = headB; else a = a.next; if(b == null) b = headA; else b = b.next; } return a; } }