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
方法一:快慢指针相遇法
注意:
- (1)fast以2倍速度绕圈和slow相遇之后回到起点
- (2)回到起点后,fast与slow同频移动直至在入口处相遇
- (3)必须考虑无环为空的情况
原理:a = c + (k-1)(b+c)
public ListNode EntryNodeOfLoop(ListNode pHead) { if(pHead == null)return null;//(1)空链表 ListNode fast = pHead; ListNode slow = pHead; while(fast != null && fast.next!=null){ fast = fast.next.next; slow = slow.next; if(slow == fast)break; } //(2)无环,所以链表结尾为空 if(fast == null || fast.next==null)return null; fast = pHead; //fast回到起点后,变慢了!按slow同样的速度前进! while(fast != slow){ fast = fast.next; slow = slow.next; } return slow; }
方法二:哈希集合法
哈希集合判定是否已存储过元素了
public ListNode EntryNodeOfLoop(ListNode pHead) { Hashset<ListNode> set = new Hashset<>(); while(cur!=null){ if(set.contains(cur)) return cur; set.add(cur); cur = cue.next; } return null; }#剑指offer#