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.

思路：假设A与B长度相等，则从链表头开始一一对比即可。

假设A与B长度不等，则进行一一对比时，短的链表会先被遍历完。这时，让该指针指向另一个链表的表头，继续进行。当长链表也被遍历完时，则其指针指向短的链表。通过这种方式，一定可以找到回合点，因为两个指针遍历的总长度是相等的，都是长链表和短链表各走了一遍。

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *ha = headA;
        ListNode *hb = headB;
        if (ha == NULL || hb == NULL) return NULL;
        while (ha != NULL && hb != NULL && ha != hb)
        {
            ha = ha->next;
            hb = hb->next;
            if (ha == hb) return ha;
            if (ha == NULL) ha = headB;
            if (hb == NULL) hb = headA;
        }
        return ha;
    }
};