java算法_删除链表的倒数第 N 个结点(LeetCode_Hot100)

问题：删除链表的倒数第 N 个结点
tip：附全部代码，复制可运行。需要代码文档，算法解析的私得

全部代码

/**
 * 2 * @Author: LJJ
 * 3 * @Date: 2023/7/13 9:17
 * 4
 */
public class LinkedListUtil {
    static class ListNode{
        int val;
        ListNode next;

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

    public static ListNode removeNthFromEnd(ListNode head,int n){

        ListNode dummy = new ListNode(0); //创造一个虚拟头头节点
        dummy.next = head;

        // 使用快慢指针定位倒数第N个节点的前一个节点
        ListNode fast = dummy;
        ListNode slow = dummy;

        dummy.next = head;

        //将快指针向前移动N+1步
        for (int i =0;i<=n;i++){
            fast = fast.next;
        }
        // 同时移动快慢指针，直到快指针到达链表尾部
        while (fast!=null){
            slow = slow.next;
            fast = fast.next;
        }
        // 删除倒数第N个节点
        slow.next = slow.next.next;

        //返回倒数第n个节点
        return dummy.next;
    }

    public static void main(String[] args) {
        // 创建链表：1 -> 2 -> 3 -> 4 -> 5
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);

        // 调用 removeNthFromEnd 方法删除倒数第2个节点
        int n = 2;
        ListNode result = removeNthFromEnd(head, n);

        // 输出链表
        while (result != null) {
            System.out.print(result.val + " -> ");
            result = result.next;
        }
        System.out.println("null");
    }
}