[Leetcode] Intersection of Two Linked Lists-优快云博客

本文链接：https://blog.youkuaiyun.com/rudyn/article/details/53375425

本文介绍了一种寻找两个单链表交汇节点的算法，通过双指针遍历的方法，在O(n)的时间复杂度和O(1)的空间复杂度下找到链表开始相交的第一个节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

题意：求得两个链表的交汇点；

思路：

1. 两个指针同时对两个链表分别遍历，当指针相同时说明存在交汇点

2. 当两个指针遍历到末尾且均不相同时说明不存在交汇点

3. 当指针A遍历至尾时，将其重定向至B链表头，继续遍历，相同地将B重定向至A，这样即便两个链表长度不一时，遍历速度快的那个指针总

能追上慢者

时间复杂度O(M+N)，空间上使用了四个临时指针

：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
		struct ListNode *pa = headA,*pb = headB;
        
        if(!pa || !pb)
        {
            return NULL;
        }
        
        struct ListNode *pa_last = NULL, *pb_last = NULL;
        
        while(pa != pb)
        {
            if(pa_last == NULL && pa->next == NULL)/* Last of PA */
            {
                pa_last = pa;
            }

            if(pb_last == NULL && pb->next == NULL)/* Last of PB */
            {
                pb_last = pb;
            }
            
            if(pa_last && pb_last && pa_last != pb_last)
            {
                return NULL;/* end of two list not equal,No intersection */
            }
            
            pa = pa->next;
            pb = pb->next;

            if(pa == NULL){
                pa = headB;
            }
            else if(pb == NULL){
                pb = headA;
            }
        }
        
        return pa;
}