首页 > 编程语言 >19. Remove Nth Node From End of List

19. Remove Nth Node From End of List

时间:2022-11-03 09:59:01浏览次数:38  
标签:Node head slow End 19 fast next start ListNode

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.

 

class Solution {

    public ListNode removeNthFromEnd(ListNode head, int n) {

        ListNode start = new ListNode(0);

        ListNode slow = start;

        ListNode fast = start;

        slow.next = head; //0的下一个结点是head

        

        for (int i = 1; i <= n + 1; i++)

            fast = fast.next; //使得slow和fast之间gap为n

        while (fast != null) {

            slow = slow.next;

            fast = fast.next;

        }

        

        slow.next = slow.next.next;

        return start.next;

    }

}

标签:Node,head,slow,End,19,fast,next,start,ListNode
From: https://www.cnblogs.com/MarkLeeBYR/p/16853419.html

相关文章