思路分析
如果没有环,非常简单,遍历一次如果到NULL那么就结束,也就证明这个链表没有环.
(这是最关键的思路,如果有环的话是见不到NULL的)
哈希表
哈希表就是遍历一次,每次遍历都把遍历过的结点放入集合中,如果下一个遍历的点存在,那么这个点就是入口结点.
cpp
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* EntryNodeOfLoop(ListNode* pHead)
{
if(!pHead)return NULL;
unordered_set<ListNode*> S;
for(auto p = pHead;p;p=p->next){
if(S.count(p)==0)
S.insert(p);
else
return p;
}
return NULL;
}
};
python
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def EntryNodeOfLoop(self, pHead):
S = set()
p = pHead
while p:
if p not in S:
S.add(p)
else: return p
p = p.next
return None
双指针
双指针的思路:
两个指针fast和slow,fast步长为2,slow步长为1.
如果fast遇到了NULL,那么证明没有环.
否则, 第一次fast和slow相遇的时候,slow在原地不动,fast回到头节点处,两指针第二次相遇的时候就是环的入口结点.
这个题难就难在证明过程,这个证明过程要熟悉.
cpp
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* EntryNodeOfLoop(ListNode* pHead)
{
auto 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; // 无环
fast = pHead;
while(fast != slow){
fast = fast->next;
slow = slow->next;
}
return fast;
}
};
python
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def EntryNodeOfLoop(self, pHead):
fast=pHead
slow=pHead
while fast and fast.next:
fast = fast.next.next
slow = slow.next
if fast==slow:break
if not fast or not fast.next:return None
fast = pHead
while fast != slow:
fast = fast.next
slow = slow.next
return fast