leetcode链表--6、linked-list-cycle-ii(有环单链表环的入口结点)

题目描述

 

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.

Follow up:

Can you solve it without using extra space?

 

解题思路：

1、确定环中结点的数目（快慢指针确定环中结点，然后遍历到下一次到该节点确定数目n）

2、一个指针先走n步，一个指针指向头结点

3、然后两个指针一起走

4、相遇点为入口点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //1、确定环中结点数目n
    //2、两个指针  1个先走n步，然后两个一起走
    //3、相遇点即为入口点
    ListNode *detectCycle(ListNode *head) {
        if(head == NULL)
            return NULL;
        ListNode *node = MettingNode(head);
        if(node == NULL)//无环
        {
            return NULL;
        }
        int nodeNum = 1;
        ListNode *pNode = node;
        while(pNode->next != node)
        {
            pNode =pNode->next;
            nodeNum++;
        }
        pNode = head;//pNode指向头结点
        for(int i=0;i<nodeNum;i++)
        {
            pNode = pNode->next;
        }
        ListNode *pNode2  = head;
        while(pNode2 != pNode)//两结点不相遇
        {
            pNode = pNode->next;
            pNode2 = pNode2->next;
        }
        return pNode;
    }
    //快慢指针找环中相遇结点
    //找到相遇节点就可确定环中结点数目
    ListNode *MettingNode(ListNode *head)
    {
        if(head == NULL)
            return NULL;
        ListNode *slow = head;
        ListNode *fast = head;
        while(fast != NULL && fast->next != NULL)
        {
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast)
                return slow;
        }
        return NULL;
    }
};