[链表]leetcode160：相交链表

题目：



题目链接：
https://leetcode-cn.com/problems/intersection-of-two-linked-lists/

解题思路：
都是利用互换遍历链表的方式来进行相遇的，第一轮指针指向短链表先走完，这就意味着它在第二轮走长的链表会慢下来，这时第一轮长链表的指针便会追上来。

代码如下：

/**
 * 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;
        }
        ListNode *last = headB;
        while (last->next != NULL) {
            last = last->next;
        }
        last->next = headB;

        ListNode *fast = headA;
        ListNode *slow = headA;

        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                slow = headA;
                while (slow != fast) {
                    slow = slow->next;
                    fast = fast->next;
                }
                last->next = NULL;
                return fast;
            }
        }
        last->next = NULL;
        return NULL;
    }    
};

标签：slow,ListNode,fast,next,链表,相交,last,NULL,leetcode160
来源： https://blog.csdn.net/qq_43152052/article/details/97421357