[leetcode 19] Remove Nth Node From End of List

1 题目

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

2 思路

题目要求遍历一遍搞定，那么，就可以用两个指针，先第一个先往前走N个，然后再一起走，最后，移除后一个元素。自己做出来的。。

3 代码

        public ListNode removeNthFromEnd(ListNode head, int n) {

            ListNode node = head;

            while(n>0 && node!=null){

                node = node.next;

                n--;

            }

            ListNode start = head;

            if(node != null){

                while(node.next != null){

                    node = node.next;

                    start = start.next;

                }

            }else{

                /* will remove the first object */

                start = start.next;

                return start;

            }

            start.next = start.next.next;

            return head;

        }

巴特西

[leetcode 19] Remove Nth Node From End of List

最新文章

热门文章