160. Intersection of Two Linked Lists

最新推荐文章于 2021-04-01 01:03:13 发布

转载最新推荐文章于 2021-04-01 01:03:13 发布 · 205 阅读

本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过使用两个指针分别遍历两个链表，当两个指针相遇时即找到了交点。此算法确保了链表结构不变，且运行时间为O(n)，空间复杂度为O(1)。

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.

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    ListNode *cur1 = headA, *cur2 = headB;
    while(cur1 != cur2){
        cur1 = cur1?cur1->next:headB;
        cur2 = cur2?cur2->next:headA;
    }
    return cur1;
}