19. 删除链表的倒数第N个节点 Remove Nth Node From End of List

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

Follow up: Could you do this in one pass?

 

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

 

方法：

     我们可以使用两个指针 first 和 second 同时对链表进行遍历， 并且first 比 second 超前 n 个节点，当 first 遍历到链表的末尾时，second 就恰好处于倒数第 n 个节点。

    如果能找到倒数第 n 个节点的前驱节点而不是倒数第 n 个节点的话，删除操作会更加方便。因此我们可以考虑在初始时将 second 指向哑节点。哑节点的next是头节点。

 

public static class ListNode {
        int val;
        ListNode next;

        ListNode() {
        }

        ListNode(int val) {
            this.val = val;
        }

        ListNode(int val, ListNode next) {
            this.val = val;
            this.next = next;
        }
    }

    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dumpy = new ListNode(0, head);
        dumpy.next = head;
        ListNode first = head;
        ListNode second = dumpy;
        for (int i = 0; i < n && first != null; i++)
            first = first.next;
        while (first != null) {
            second = second.next;
            first = first.next;
        }
        if (second.next != null)
            second.next = second.next.next;
        else
            second.next = null;
        return dumpy.next;
    }

 

参考链接：

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list