[Leetcode] 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.

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;}