给一个长度为n链表,若其中包含环,请找出该链表的环的入口结点,否则,返回null。
数据范围:
,
要求:空间复杂度
,时间复杂度
例如,输入{1,2},{3,4,5}时,对应的环形链表如下图所示:
可以看到环的入口结点的结点值为3,所以返回结点值为3的结点。
输入分为2段,第一段是入环前的链表部分,第二段是链表环的部分,后台会根据第二段是否为空将这两段组装成一个无环或者有环单链表
返回链表的环的入口结点即可,我们后台程序会打印这个结点对应的结点值;若没有,则返回对应编程语言的空结点即可。
{1,2},{3,4,5}3
返回环形链表入口结点,我们后台程序会打印该环形链表入口结点对应的结点值,即3
{1},{}"null"
没有环,返回对应编程语言的空结点,后台程序会打印"null"
{},{2}
2
环的部分只有一个结点,所以返回该环形链表入口结点,后台程序打印该结点对应的结点值,即2
function EntryNodeOfLoop(pHead)
{
// write code here
if (pHead.next === null) return null
const map = new Map()
let currentNode = pHead
while(currentNode) {
if (map.has(currentNode)) return currentNode
else map.set(currentNode, currentNode)
currentNode = currentNode.next
}
return null
} /*function ListNode(x){
this.val = x;
this.next = null;
}*/
function EntryNodeOfLoop(pHead)
{
// write code here
if(!pHead){
return null;
}
let key=0;
let obj={};
while(pHead){
key = pHead.val;//节点值
if(obj[key]){//已有节点,则有环,返回节点
return pHead;
}else{//没有,继续执行,数量为1
obj[key]=1;
pHead = pHead.next;
}
}
}
module.exports = {
EntryNodeOfLoop : EntryNodeOfLoop
}; function EntryNodeOfLoop(pHead)
{
while(pHead !== null){
if(pHead.flag){
return pHead
}else{
pHead.flag = true;
pHead = pHead.next
}
}
return null
} function EntryNodeOfLoop(pHead)
{
// write code here
let fast =pHead;
let slow =pHead;
while(fast&&fast.next){
fast=fast.next.next;
slow=slow.next;
if(fast==slow) break;
};
if(!fast||!fast.next) return null
fast=pHead;
while(fast!=slow){
fast=fast.next;
slow=slow.next;
};
return fast
} function EntryNodeOfLoop(pHead)
{
// write code here
if(!pHead) return null
let fast = pHead, slow = pHead
while(fast && fast.next){
fast = fast.next.next
slow = slow.next
if(fast == slow){
while(pHead !== slow){
pHead = pHead.next
slow = slow.next
}
return slow
}
}
return null
} function EntryNodeOfLoop(pHead)
{
// write code here
var fast = pHead,
slow = pHead;
while(fast && fast.next){
fast = fast.next.next;
slow = slow.next;
if(fast == slow){
break;
}
}
if(!fast || !fast.next){
return null;
}
fast = pHead;
while(fast != slow){
fast = fast.next;
slow = slow.next;
}
return fast;
} function EntryNodeOfLoop(pHead)
{
// write code here
if(!pHead){
return null
}
let temp = pHead
let arr = []
while(temp){
if(arr.includes(temp)){
return temp
}
arr.push(temp)
temp = temp.next
}
return null
} /*function ListNode(x){
this.val = x;
this.next = null;
}*/
function EntryNodeOfLoop(pHead)
{
// write code here
let fast = pHead;
let slow = pHead;
while(fast && fast.next){
fast = fast.next.next;
slow = slow.next;
if(fast == slow){
fast = pHead;
while(fast != slow){
fast = fast.next;
slow = slow.next;
}
return fast
}
}
return null;
} /*function ListNode(x){
this.val = x;
this.next = null;
}*/
function EntryNodeOfLoop(pHead)
{
// write code here
if(!pHead) return null;
let nodeList = [];
let currNode = pHead;
while(currNode !== null){
if(nodeList.indexOf(currNode) !== -1){
return currNode;
}
nodeList.push(currNode);
currNode = currNode.next;
}
return null;
} /*function ListNode(x){
this.val = x;
this.next = null;
}*/
function EntryNodeOfLoop(pHead)
{
// write code here
let tmp = {}
let p = pHead
while(p) {
if(!tmp[p.val]) {
tmp[p.val] = 1
p = p.next
}else {
return p
}
}
return null
} function EntryNodeOfLoop(pHead)
{
let h = pHead;
if(!h.next || !h.next.next){
return null;
}
let fast = h.next.next;
let slow = h.next;
while(fast!==null && fast.next!==null){
if(fast === slow){
fast = h;
while(fast !==slow){
fast=fast.next;
slow=slow.next;
}
return slow;
}else{
fast = fast.next.next;
slow = slow.next;
}
}
return null;
} function EntryNodeOfLoop(pHead) {
if (!pHead || !pHead.next) {
return null;
}
let P1 = pHead.next;
let P2 = pHead.next.next;
// 1.判断是否有环
while (P1 != P2) {
if (P2 === null || P2.next === null) {
return null;
}
P1 = P1.next;
P2 = P2.next.next;
}
// 2.获取环的长度
let temp = P1;
let length = 1;
P1 = P1.next;
while (temp != P1) {
P1 = P1.next;
length++;
}
// 3.找公共节点
P1 = P2 = pHead;
while (length-- > 0) {
P2 = P2.next;
}
while (P1 != P2) {
P1 = P1.next;
P2 = P2.next;
}
return P1;
}
// 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。
/*function ListNode(x){
this.val = x;
this.next = null;
}*/
// 解题思路:
// 1. 一个指针走一步,一个指针走两步,如果在到Null之前重合,则证明有环存在
// 2. 假设这个环走了若干个结点,但是由于要圈上 多走整数圈数 才能相遇,则这个等式成立 2x = x + r*n,r为圈数
// 3. 那么此时新建一个从头开始,走一步的指针,同时继续走,走x步即可回到相遇点,但是如果在相遇点之前重合,则重合点为入口
function EntryNodeOfLoop(pHead)
{
if(!pHead || !pHead.next) {
return null
}
let fast = pHead
let slow = pHead
// 题目已知有环
while(1) {
fast = fast.next
slow = slow.next.next
if(fast === slow) {
break
}
}
let result = pHead
while(result !== slow) {
slow = slow.next
result = result.next
}
return result
}
当第二个指针指向环的入口结点时,第一个指针已经围绕着环走了一圈又回到了入口结点。
public class 链表中环的入口结点 { //找到一快一满指针相遇处的节点,相遇的节点一定是在环中 public static ListNode meetingNode(ListNode head) { if(head==null) return null; ListNode slow = head.next; if(slow==null) return null; ListNode fast = slow.next; while (slow != null && fast != null) { if(slow==fast){ return fast; } slow=slow.next; fast=fast.next; if(fast!=slow){ fast=fast.next; } } return null; } public ListNode EntryNodeOfLoop(ListNode pHead) { ListNode meetingNode=meetingNode(pHead); if(meetingNode==null) return null; // 得到环中的节点个数 int nodesInLoop=1; ListNode p1=meetingNode; while(p1.next!=meetingNode){ p1=p1.next; ++nodesInLoop; } // 移动p1 p1=pHead; for(int i=0;i<nodesInLoop;i++){ p1=p1.next; } // 移动p1,p2 ListNode p2=pHead; while(p1!=p2){ p1=p1.next; p2=p2.next; } return p1; } }