1 
public class Solution {

  /**
   * Return a list containing, at most, the first numNodes nodes in the list starting with head. If
   * numNodes is larger than the length of the list, then the entire list will be returned.
   * 
   * Examples:
   * 
   * <pre>
   * [11, 12, 13], numNodes = 2
   * returns [11, 12]
   * 
   * [11, 12, 13], numNodes = 0
   * returns []
   * 
   * [11, 12, 13], numNodes = 100
   * returns [11, 12, 13]
   * 
   * [], numNodes = 5
   * returns []
   * </pre>
   * 
   * @param head - Head of the list to truncate
   * @param numNodes - Maximum number of nodes in the resulting list
   * @return head of the truncated list
   */
  public static ListNode truncate(ListNode head, int numNodes) {
    // just return null
    if (head == null) {
      return null;
    }
    // Recursion head.next
    if (head.next != null) {
      head.next = truncate(head.next, numNodes--);
    }
    
    if (numNodes <= 0) {
      return head.next;
    } else {
      return head;
    }
  }

如您所见，此代码的目的是截断超过 numNodes 的每个节点。除了通过长度为 4 的列表并且我需要返回前 3 个节点（numNodes = 3）的情况之外，我还通过了所有测试。如果有人可以帮助我，那将不胜感激。我不知道我做错了什么。

这是 ListNode 类

/**
 * Simple Singly-linked-list node.
 */
public class ListNode {

  public int val;
  public ListNode next;

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

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

  public ListNode() {
    this.val = 0;
    this.next = null;
  }
}

这是我唯一失败的测试

testTruncateLengthFourListToThree() 预期：[10,11,12] 实际：[10,11,12,13]

解决方案必须是递归的。

4

2 回答 2

0

尝试这个。

record ListNode(int val, ListNode next) {}

public static ListNode truncate(ListNode head, int numNodes) {
    if (head == null || numNodes <= 0)
        return null;
    else
        return new ListNode(head.val, truncate(head.next, numNodes - 1));
}

public static String toString(ListNode head) {
    List<String> list = new ArrayList<>();
    for ( ; head != null; head = head.next)
        list.add("" + head.val);
    return list.toString();
}

public static void main(String[] args) {
    ListNode list = new ListNode(1, new ListNode(2, new ListNode(3, null)));
    System.out.println("list = " + toString(list));
    System.out.println("truncate(list, 3) = " + toString(truncate(list, 3)));
    System.out.println("truncate(list, 2) = " + toString(truncate(list, 2)));
    System.out.println("truncate(list, 1) = " + toString(truncate(list, 1)));
    System.out.println("truncate(list, 0) = " + toString(truncate(list, 0)));
}

输出：

list = [1, 2, 3]
truncate(list, 3) = [1, 2, 3]
truncate(list, 2) = [1, 2]
truncate(list, 1) = [1]
truncate(list, 0) = []
于 2021-10-13T01:08:58.120 回答
0

也许您可以在递归调用 truncate 时使用前缀增量而不是后缀，从 truncate(head.next, numNodes--) 更改为 truncate(head.next, --numNodes) 以便在递归之前减少 numNodes。

于 2021-10-13T01:14:56.707 回答