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.
首先找到两条链表的长度, 然后截取相同的长度, 从两个开端一起走 碰到一起的话返回碰到时候的node, 没有相遇的话返回null.
public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { int lena = getlen(headA); int lenb = getlen(headB); ListNode nodea = headA; ListNode nodeb = headB; if (lena > lenb) { for (int i = 0; i < lena - lenb; i++) { nodea = nodea.next; } } else { for (int i = 0; i < lenb - lena; i++) { nodeb = nodeb.next; } } while (nodea != null && nodeb!= null) { if (nodea == nodeb) { return nodea; } nodea = nodea.next; nodeb = nodeb.next; } return null; } public int getlen(ListNode head) { int count = 0; while (head != null) { head = head.next; count++; } return count; } }
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