LeetCode—Intersection of Two Linked Lists 找两个链表的相交位置，让长的链表先走一段

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. 只能够是O（n）的时间复杂度，这里的做法是首先遍历两个链表知道长度，然后让长的链表先走一段，然后再进行两两比较，时间复杂度O（lengthA+lengthB+maxLength）就是O（n）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA == NULL || headB == NULL)
        {
            return NULL;
        }
        int lengthA = 1;
        int lengthB = 1;
        ListNode *Afirst = headA;
        ListNode *Bfirst = headB;
        while(Afirst->next)
        {
            lengthA++;
            Afirst = Afirst->next;
        }
        while(Bfirst->next)
        {
            lengthB++;
            Bfirst = Bfirst->next;
        }
        if(Afirst != Bfirst)
        {
            return NULL;
        }
        int length = lengthA;//<随便初始化一个
        if(lengthA > lengthB)
        {
            int diff = lengthA-lengthB;
            length = lengthB;
            while(diff--)
            {
                headA = headA->next;
            }
        }
        if(lengthA < lengthB)
        {
            length = lengthA;
            int diff = lengthB-lengthA;
            while(diff--)
            {
                headB = headB->next;
            }
        }
        while(length--)
        {
            if(headA == headB)
            {
                return headA;
            }
            headA = headA->next;
            headB = headB->next;
        }
        return NULL;

    }
};