剑指offer（52）-链表中环的入口结点

题目描述

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一个链表中包含环，请找出该链表的环的入口结点。

代码

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/*struct ListNode {    int val;    struct ListNode *next;    ListNode(int x) :        val(x), next(NULL) {    }};*/class Solution {public:    //找到一快一慢两个指针相遇的结点，两个指针相遇的结点一定在环中。    ListNode* MeetingNode(ListNode* pHead)    {        if (pHead == NULL) {            return NULL;        }        ListNode *pSlow = pHead ->next;        if (pSlow == NULL) {            return NULL;        }        ListNode *pFast = pHead->next->next;        while (pFast != NULL && pSlow != NULL) {            if (pFast == pSlow) {                return pFast;            }            pSlow = pSlow ->next;            pFast = pFast ->next;            if (pFast != NULL) {                pFast = pFast ->next;            }        }        return NULL;    }    // 在找到环中任意一个结点之后，就能得出环中的结点数目，并找到环中的入口结点。    ListNode* EntryNodeOfLoop(ListNode* pHead)    {        ListNode *meetingNode = MeetingNode(pHead);        if (meetingNode == NULL) {            return NULL;        }        // get the number of nodes in loop        int nodesInLoop = 1;        ListNode *pNode1 = meetingNode;        while (pNode1->next != meetingNode) {            pNode1 = pNode1 ->next;            nodesInLoop++;        }        // move pNode1        pNode1 = pHead;        for (int i = 0; i < nodesInLoop; i++) {            pNode1 = pNode1 ->next;        }        // move pNode1 and pNode2        ListNode *pNode2 = pHead;        while (pNode1 != pNode2) {            pNode1 = pNode1 ->next;            pNode2 = pNode2 ->next;        }        return pNode1;    }};