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
public class Solution {
// 思路：
// 1、判断链表是否有环
// 使用快慢指针，快指针走两步，慢指针走一步，如果快慢指针相遇就有环（循环条件快指针和快指针的next不为空）
// 2、找相遇节点
// 返回找环时相遇的慢指针，定义另一个快指针从头出发，快慢指针相遇处就是入口
public ListNode EntryNodeOfLoop(ListNode pHead) {
ListNode slow = judgeCircle(pHead);
if (slow == null) return null;
ListNode fast = pHead;
while (fast != slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
public ListNode judgeCircle(ListNode head) {
ListNode slow = head,quick = head;
while (quick != null && quick.next != null) {
slow = slow.next;
quick = quick.next.next;
if (slow == quick) return slow;
}
return null;
}
}
