19. 删除链表的倒数第N个节点

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19. 删除链表的倒数第N个节点

官方链接

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

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。

进阶：

你能尝试使用一趟扫描实现吗？

解法一

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {

        if (head == null) return head;

        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode slow = dummy, fast = dummy;

        while(n-- > 0) fast = fast.next;

        while(fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;

        return dummy.next;
    }
}

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        slow = fast = dummy
        for i in range(n):
            if fast and fast.next:
                fast = fast.next
        while fast.next:
            slow = slow.next
            fast = fast.next

        slow.next = slow.next.next
        return dummy.next

解法二 无哨兵

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null) return head;

        ListNode slow = head, fast = head;
        while(n-- > 0) fast = fast.next;
        if(fast == null) return head.next;

        while(fast.next != null){
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return head;
    }
}