Intersection of Two Linked Lists

最新推荐文章于 2024-06-24 11:25:34 发布

原创最新推荐文章于 2024-06-24 11:25:34 发布 · 256 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

294 篇文章

订阅专栏

本文介绍了一种在两个单链表中寻找它们开始相交节点的方法。通过两次遍历链表，首先确定两个链表的长度，然后使较长的链表先前进一定步数，再同时遍历两个链表直至找到相同的节点。

https://oj.leetcode.com/problems/intersection-of-two-linked-lists/

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.

public ListNode getIntersectionNode(ListNode headA, ListNode headB)

这题只要求O(N)和O(1)，但没有要求一遍过。所以两个链表各扫两遍。第一遍扫出各自长度。然后让长的先行，当长的走过了长度差的节点之后。相当于现在长链表和短链表剩余长度一样，然后再一起前行。直到找到长短链表指针都相等或者走完为止。

给出代码如下：

    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int lengthA = 0;
        int lengthB = 0;
        ListNode tmpA = headA;
        while(tmpA != null){
            tmpA = tmpA.next;
            lengthA++;
        }
        ListNode tmpB = headB;
        while(tmpB != null){
            tmpB = tmpB.next;
            lengthB++;
        }
        tmpA = headA;
        tmpB = headB;
        while(tmpA != null && tmpB != null){
            if(tmpA == tmpB)
                return tmpA;
            if(lengthA > lengthB){
                tmpA = tmpA.next;
                lengthA--;
            }else if(lengthA < lengthB){
                tmpB = tmpB.next;
                lengthB--;
            }else{
                tmpA = tmpA.next;
                tmpB = tmpB.next;
            }
        }
        return null;
    }