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 = 0; i < lenA - lenB; ++i) pA = pA->next;
            else for(int i = 0; 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 = 0; i < lenA - lenB; ++i) pA = pA->next;
                else for(int i = 0; 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 = 0;
        while(head != end) {
            ++ret;
            head = head->next;
        }
        return ret;
    }
};