Intersection of Two Linked Lists 查找2个链表的公共交点

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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; *///查找两个链表的第一个公共节点，如果两个节点的尾节点相同，肯定存在公共节点//长的链表开始多走 （h1的数量 - h2的数量）步，然后和短链表同步往下走，遇到的第一个相同的节点就是最早的公共节点class Solution {public:    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {        if(headA==NULL||headB==NULL)            return NULL;        ListNode *p=headA;        ListNode *q=headB;        int k1=1,k2=1;        while(p->next)        {            p=p->next;            k1++;        }        while(q->next)        {            q=q->next;            k2++;        }        if(p!=q)            return NULL;        else        {            int count=abs(k2-k1);            if(k2>k1)            {                p=headB;                q=headA;            }            else            {                p=headA;                q=headB;            }            while(count--)            {                p=p->next;            }            while(p&&q&&p!=q)            {                p=p->next;                q=q->next;            }            return p;        }    }};