LeetCode 相交链表&环形链表II

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

题目连接：https://leetcode-cn.com/problems/linked-list-cycle-ii/

题目大意

　　具体按左神书上的描述来实现，空间复杂度 O(1) 解决有环单链表相交问题。

分析

　　关于快慢指针能找到入环节点的一些说明强烈推荐这篇博客：https://blog.csdn.net/ffj0721/article/details/83302412

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) {
         ListNode *loopA = detectCycle(headA), *loopB = detectCycle(headB);
         int lenA, lenB;

         if(loopA == NULL && loopB == NULL) { // 两个链表都没有环
             lenA = getLen(headA);
             lenB = getLen(headB);

             // 长链表先走，然后再一起走
             ListNode *pA = headA, *pB = headB;
             if(lenA > lenB) for(int i = ; i < lenA - lenB; ++i) pA = pA->next;
             else for(int i = ; i < lenB - lenA; ++i) pB = pB->next;

             while(pA != pB) {
                 pA = pA->next;
                 pB = pB->next;
             }

             return pA;
         }
         else if(loopA != NULL && loopB != NULL) { // 两个链表都有环
             if(loopA == loopB) { // 和都没有环的情况一样，只不过终点不是NULL了
                 lenA = getLen(headA, loopA);
                 lenB = getLen(headB, loopB);

                 ListNode *pA = headA, *pB = headB;
                 if(lenA > lenB) for(int i = ; i < lenA - lenB; ++i) pA = pA->next;
                 else for(int i = ; i < lenB - lenA; ++i) pB = pB->next;

                 while(pA != pB) {
                     pA = pA->next;
                     pB = pB->next;
                 }

                 return pA;
             }
             else {
                 ListNode *p = loopA->next;

                 while(p != loopA) {
                     if(p == loopB) return loopA; // 相交但入环点不同
                     p = p->next;
                 }
                 return NULL; // 不相交
             }
         }
         return NULL; // 一个有环一个没环，一定不相交
     }

     // 如果链表无环，返回 NULL，否则返回第一个入环节点
     ListNode *detectCycle(ListNode *head) {
         if(head == NULL || head->next == NULL || head->next->next == NULL) return NULL;

         ListNode *slow = head->next, *fast = head->next->next;

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

         slow = head;
         while(fast != slow) {
             fast = fast->next;
             slow = slow->next;
         }

         return slow;
     }

     int getLen(ListNode *head, ListNode *end = NULL) {
         int ret = ;
         while(head != end) {
             ++ret;
             head = head->next;
         }
         return ret;
     }
 };