题目描述：

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.

思路分析：先求出两段链表的长度length_a、length_b，让较长链表的头指针先移动|length_a-length_b|个单位。然后让两个链表的头指针一起移动，当两个指向两个链表的指针相等时，即找到了交点。

代码：

ListNode * Solution::getIntersectionNode(ListNode *headA, ListNode *headB)
{
    ListNode * a = headA;
    ListNode * b = headB;
    int length_a = 0,length_b = 0;
    while(a != NULL)
    {
        length_a++;
        a = a->next;
    }
    while(b != NULL)
    {
        length_b++;
        b = b->next;
    }
    if(length_a > length_b)
    {
        int difference = length_a - length_b;
        while(difference > 0)
        {
            difference--;
            headA = headA->next;
        }

    }
    else if(length_b > length_a)
    {
        int difference = length_b - length_a;
        while(difference > 0)
        {
            difference--;
            headB = headB->next;
        }
    }
    while(headA != headB)
    {
        headA = headA->next;
        headB = headB->next;
    }
    return headA;
}

Intersection of Two Linked Lists