给
定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
示例:
给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5.
说明:
给定的 n 保证是有效的。
/** * Definition for singly-linked list. * public class ListNode { * public int val; * public ListNode next; * public ListNode(int x) { val = x; } * } */ public class Solution { public ListNode RemoveNthFromEnd(ListNode head, int n) { ListNode p = head; int count = 0; while(p!=null){ count++; p = p.next; } p = head; if(count < n){ return null; } if(count == n){ return head.next; } for(int i=0; i<count-n-1; i++){ if(p.next != null){ p = p.next; } } if(p.next != null){ p.next = p.next.next; } return head; } }
/** * Definition for singly-linked list. * public class ListNode { * public int val; * public ListNode next; * public ListNode(int x) { val = x; } * } */ public class Solution { public ListNode RemoveNthFromEnd(ListNode head, int n) { ListNode p = head; List<ListNode> nodes = new List<ListNode>(); while(p != null){ nodes.Add(p); p = p.next; } if(n == nodes.Count){ return head.next; } nodes[nodes.Count - n -1].next = nodes[nodes.Count - n -1].next.next; return head; } }