160 intersection of two linked list

两个指针分别单步遍历链表，当访问到尾部额时候，指向另外一个链表的头继续遍历，直到两个指针相遇就返回

对于[1] 两个都指向[1] ，一开始就出现p1 == p2情况，需要直接返回p1或者p2

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *p1 = headA;
    struct ListNode *p2 = headB;

    if (p1 == NULL || p2 == NULL) return NULL;

    while (p1 != NULL && p2 != NULL && p1 != p2) {
        p1 = p1->next;
        p2 = p2->next;
        //
        // Any time they collide or reach end together without colliding 
        // then return any one of the pointers.
        //
        if (p1 == p2) return p1;
        //
        // If one of them reaches the end earlier then reuse it 
        // by moving it to the beginning of other list.
        // Once both of them go through reassigning, 
        // they will be equidistant from the collision point.
        //
        if (p1 == NULL) p1 = headB;
        if (p2 == NULL) p2 = headA;
    }
    return p1;    
}