每天一道算法题（java数据结构与算法)——＞删除链表的倒数第 N 个结点

Posted 2022-01-07 stormzhuo

题目
给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例1

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例2

输入：head = [1], n = 1
输出：[]

示例2

输入：head = [1,2], n = 1
输出：[1]

题解

1. 定义两个指针front,back指向头结点

2. 第1个指针front从链表的头结点开始遍历k步，第2个指针back保持不动；

3. 从第k+1步开始指针back也从链表的头结点开始和指针front以相同的速度遍历。

4. 当指针p1指向链表的尾节点时指针p2正好指向倒数第k+1个节点(两个指针的距离始终保持为k)

5. 把倒数第k+1个节点的next指针指向倒数第k-1个节点

代码实现

解类如下，类里包含两个方法，一个是遍历链表的，一个是删除链表的倒数第 N 个结点的

public class Solution 

    public static ListNode removeNthFromEnd(ListNode head, int n) 
        // 定义一个哨兵结点作为虚拟头结点,这样可以减少判断头结点是否为null和删除头结点逻辑判断
        ListNode<Integer> dummy = new ListNode<>(0);
        // 让哨兵结点指向头结点
        dummy.next = head;
        // 定义两个结点初始化为哨兵结点
        ListNode front = dummy;
        ListNode back = dummy;
        // front遍历n步
        for (int i = 0; i < n; i++) 
            front = front.next;
        
         /* front,back以相同的速度遍历，front接上次位置遍历，back从哨兵遍历
          循环终止条件为front到达尾结点时*/
        while (front.next != null) 
            front = front.next;
            back = back.next;
        
        // 此时back的下一个结点就是要删除的结点，让back指向下一个结点的下一个结点
        back.next = back.next.next;
        return dummy.next;
    

    public static void printList(ListNode head) 
        System.out.print("[");
        while (head != null) 
            if (head.next == null) 
                System.out.print(head.val);
             else 
                System.out.print(head.val + ",");
            
            head = head.next;
        
        System.out.print("]");
    

结点类

public class ListNode<T> 

    T val;
    ListNode next;

    public ListNode() 

    

    public ListNode(T val) 
        this.val = val;
    

    public ListNode(T val, ListNode next) 
        this.next = next;
    

测试类

public class Test 

    @org.junit.Test
    public void test1() 
        ListNode<Integer> head = new ListNode<>(1);
        ListNode<Integer> listNode = new ListNode<>(2);
        ListNode<Integer> listNode1 = new ListNode<>(3);
        ListNode<Integer> listNode2 = new ListNode<>(4);
        ListNode<Integer> listNode3 = new ListNode<>(5);
        head.next = listNode;
        listNode.next = listNode1;
        listNode1.next = listNode2;
        listNode2.next = listNode3;
        System.out.print("输入：");
        Solution.printList(head);
        System.out.println(" 2");
        ListNode head2 = Solution.removeNthFromEnd(head, 2);
        System.out.print("输出:");
        Solution.printList(head2);
    

    @org.junit.Test
    public void test2() 
        ListNode<Integer> head = new ListNode<>(1);
        System.out.print("输入：");
        Solution.printList(head);
        System.out.println(" 1");
        ListNode head2 = Solution.removeNthFromEnd(head, 1);
        System.out.print("输出:");
        Solution.printList(head2);
    

    @org.junit.Test
    public void test3() 
        ListNode<Integer> head = new ListNode<>(1);
        ListNode<Integer> listNode = new ListNode<>(2);
        head.next = listNode;
        System.out.print("输入：");
        Solution.printList(head);
        System.out.println(" 2");
        ListNode head2 = Solution.removeNthFromEnd(head, 1);
        System.out.print("输出:");
        Solution.printList(head2);
    

测试结果