课后作业
问题描述
原文链接:142. 环形链表 II
给定一个链表的头节点 head
,返回链表开始入环的第一个节点。 如果链表无环,则返回 null
。
如果链表中有某个节点,可以通过连续跟踪 next
指针再次到达,则链表中存在环。 为了表示给定链表中的环,评测系统内部使用整数 pos
来表示链表尾连接到链表中的位置(索引从 0 开始)。如果 pos
是 -1
,则在该链表中没有环。注意:pos
不作为参数进行传递,仅仅是为了标识链表的实际情况。
不允许修改链表。
示例 1:
输入:head = [3,2,0,-4], pos = 1
输出:返回索引为 1 的链表节点
解释:链表中有一个环,其尾部连接到第二个节点。
示例 2:
输入:head = [1,2], pos = 0
输出:返回索引为 0 的链表节点
解释:链表中有一个环,其尾部连接到第一个节点。
示例 3:
输入:head = [1], pos = -1
输出:返回 null
解释:链表中没有环。
提示:
- 链表中节点的数目范围在范围
[0, 104]
内 -105 <= Node.val <= 105
pos
的值为-1
或者链表中的一个有效索引
代码实现
Java
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head == null || head.next == head){
return head;
}
//
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null){
fast = fast.next.next;
slow = slow.next;
if(fast == slow){
break;
}
}
if(fast == null || fast.next == null){
return null;
}
//统计环的节点个数
int n = 1;
while(true){
fast = fast.next;
if(fast == slow){
break;
}
n++;
}
ListNode pre = head;
ListNode last = head;
// last先走n步
for(int i = 0; i < n; i++){
last = last.next;
}
while(true){
if(last == pre){
return pre;
}
last = last.next;
pre = pre.next;
}
}
}
Python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if(head == None or head.next == head):
return head
#
fast = head
slow = head
while(fast != None and fast.next != None):
fast = fast.next.next
slow = slow.next
if(fast == slow):
break
if(fast == None or fast.next == None):
return None
#统计环的节点个数
n = 1
while(True):
fast = fast.next
if(fast == slow):
break
n += 1
pre = head
last = head
# last先走n步
for i in range(n):
last = last.next
while(True):
if(last == pre):
return pre
last = last.next
pre = pre.next
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
if(head == NULL || head->next == head){
return head;
}
//
ListNode* fast = head;
ListNode* slow = head;
while(fast != NULL && fast->next != NULL){
fast = fast->next->next;
slow = slow->next;
if(fast == slow){
break;
}
}
if(fast == NULL || fast->next == NULL){
return NULL;
}
//统计环的节点个数
int n = 1;
while(true){
fast = fast->next;
if(fast == slow){
break;
}
n++;
}
ListNode* pre = head;
ListNode* last = head;
// last先走n步
for(int i = 0; i < n; i++){
last = last->next;
}
while(true){
if(last == pre){
return pre;
}
last = last->next;
pre = pre->next;
}
}
};
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func detectCycle(head *ListNode) *ListNode {
if head == nil || head.Next == head {
return head
}
//
fast := head
slow := head
for fast != nil && fast.Next != nil {
fast = fast.Next.Next
slow = slow.Next
if fast == slow {
break
}
}
if fast == nil || fast.Next == nil {
return nil
}
//统计环的节点个数
n := 1
for {
fast = fast.Next
if fast == slow {
break
}
n++
}
pre := head
last := head
// last先走n步
for i := 0; i < n; i++ {
last = last.Next
}
for {
if last == pre {
return pre
}
last = last.Next
pre = pre.Next
}
}