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:
/** * 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 curA = headA,curB = headB; while (curA!=null&&curB!=null){ if (curA.next ==null&&curB.next==null&&curA!=curB) return null; //如果走到了链的尽头,而这尽头又不是交 //点,说明两条链不相交 else if (curA==curB) //如果同时到交点了,就跳出循环 break; if (curA.next == null) //下面这段可能需要画图理解 curA = headB; //如果两条链有交点,则curA和curB到 else //交点的距离是相同的 curA = curA.next; if (curB.next == null) curB = headA; else curB = curB.next; } return curA; } }
- 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.