技術標籤：leetcode刷題leetcode連結串列

刪除連結串列的倒數第N個節點

• 做題部落格連結
• 題目連結
• 描述
• 示例
• 初始程式碼模板
• 程式碼

做題部落格連結

https://blog.csdn.net/qq_43349112/article/details/108542248

題目連結

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

描述

給你一個連結串列，刪除連結串列的倒數第 n 個結點，並且返回連結串列的頭結點。

進階：你能嘗試使用一趟掃描實現嗎？

提示：

連結串列中結點的數目為 sz
1 <= sz <=
 
 30
0 <= Node.val <= 100
1 <= n <= sz

示例

示例 1：

輸入：head = [1,2,3,4,5], n = 2
輸出：[1,2,3,5]

示例 2：

輸入：head = [1], n = 1
輸出：[]

示例 3：

輸入：head = [1,2], n = 1
輸出：[1]

初始程式碼模板

/**
 * 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) {
       
    }
}

程式碼

/**
 * 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) {
        ListNode prevHead = new ListNode(-1);
        prevHead.next = head;

        ListNode fast = prevHead;
        ListNode slow = prevHead;

        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }

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

        return prevHead.next;
    }
}