55链表中环的入口结点
题目描述
一个链表中包含环，请找出该链表的环的入口结点。
利用快慢指针
如果链表的环中有n个节点。快指针先走n步，然后同时走。当慢指针走的入口节点的时候，
快指针围绕着环走了一圈又回到了入口节点。
如何得到得到环中节点的个数？
利用快慢指针，找到环中的任意一节点，从该节点走一圈，并计数，当再次回到该节点的时候，便统计完了环中节点的个数。
1 /* 2 public class ListNode { 3 int val; 4 ListNode next = null; 5 6 ListNode(int val) { 7 this.val = val; 8 } 9 } 10 */ 11 public class Solution { 12 public ListNode EntryNodeOfLoop(ListNode head) 13 { 14 ListNode pMeet = MeetingNode(head); 15 if(pMeet==null) return null; 16 17 ListNode fast = head.next; 18 ListNode slow = head; 19 for(ListNode p = pMeet.next;p!=pMeet;p=p.next) 20 fast = fast.next; 21 22 while(slow!=fast){ 23 slow = slow.next; 24 fast = fast.next; 25 26 } 27 return fast; 28 } 29 30 31 32 private ListNode MeetingNode(ListNode head){ 33 if(head ==null|| head.next==null) return null; 34 ListNode slow = head; 35 ListNode fast = head.next.next; 36 while(slow!=fast){ 37 slow = slow.next; 38 fast = fast.next.next; 39 } 40 return slow; 41 } 42 43 44 45 46 }
c++:20180729
1 /* 2 struct ListNode { 3 int val; 4 struct ListNode *next; 5 ListNode(int x) : 6 val(x), next(NULL) { 7 } 8 }; 9 */ 10 class Solution { 11 public: 12 ListNode* EntryNodeOfLoop(ListNode* head) 13 { 14 15 ListNode* meetNode= meeting(head); 16 if(meetNode==NULL) return NULL; 17 ListNode* fast = head; 18 ListNode* slow = head; 19 for(ListNode* p = meetNode;p!=meetNode;p=p->next) 20 fast=fast->next; 21 22 while(slow!=fast){ 23 fast = fast->next; 24 slow = slow->next; 25 } 26 return fast; 27 } 28 29 ListNode* meeting(ListNode* head){ 30 if(head==NULL ||head->next==NULL) return NULL; 31 ListNode* fast = head->next->next; 32 ListNode* slow = head; 33 while(slow!=fast){ 34 fast = fast->next->next; 35 slow = slow->next; 36 } 37 return fast; 38 } 39 };
