【重点】19.删除链表的倒数第N个结点-优快云博客

本文链接：https://blog.youkuaiyun.com/Allenlzcoder/article/details/134923469

文章介绍了在链表操作中，如何使用头节点和不使用头节点的方法来删除给定位置的节点，通过快慢指针实现，包括法1（使用虚拟头结点）和法2（直接操作）。

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Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = ListNode(-1)
        dummy.next = head
        slow = dummy # slow, fast初始位置都在虚拟头结点上
        fast = dummy
        for i in range(0, n):
            fast = fast.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy.next

Java

法1：使用头结点

注意使用虚拟头结点！！！

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null || head.next == null) {
            return null;
        }
        ListNode dummy = new ListNode(-1); // 虚拟头结点
        dummy.next = head;
        ListNode slow = dummy, fast = dummy;
        while (n > 0) {
            fast = fast.next;
            --n;
        }
        while (fast != null && fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}

法2：不使用头结点

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) {
            return null;
        }
        ListNode fast = head, slow = head;
        while (n > 0 && fast != null) {
            fast = fast.next;
            --n;
        }
        if (fast == null) { // 删除正数第1个结点
            return head.next;
        }

        while (fast.next != null) { // fast停留在最后的结点上, slow停留在被删除的前一个结点上
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;

        return head;
    }
}