题解 | #链表中环的入口结点#
链表中环的入口结点
https://www.nowcoder.com/practice/253d2c59ec3e4bc68da16833f79a38e4
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* EntryNodeOfLoop(ListNode* pHead) {
ListNode *pFast=pHead, *pSlow = pHead;
while (pFast != nullptr)
{
if (pFast->next == nullptr)
return nullptr;
pSlow = pSlow->next; // 慢指针一次一步
pFast = pFast->next->next; // 快指针一次2步
if(pSlow == pFast) // 两者相遇
{
pFast = pHead; // 快指针回到链表头
while (pFast != pSlow)
{
pFast = pFast->next; // 快慢指针都是一步一步走,直到再次相遇,即为环入口
pSlow = pSlow->next;
}
return pFast;
}
}
return nullptr;
}
};
