[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;
}