我正在尝试一个leetcode问题,我需要在java中实现一个链表,但是“链接”永远不会被创建。节点本身确实会被创建,但会丢失在内存中。我知道如何在c++中使用指针来实现这一点,但是在java中如何实现呢?
问题是:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
印刷品:
7
0
8
返回:
7 (just head node)
我的代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//hold root node to return later, use temp node (l3) to create list
ListNode head = new ListNode(0);
ListNode l3 = head;
boolean carryover = false;
//if lists l1, l2 still have a value, append to l3
while (l1 != null || l2 != null)
{
//always true except on first iteration
if (l3 == null)
l3 = new ListNode(0);
//if l1.val + l2.val >= 10 from last iteration, carry over 1
if (carryover)
{
l3.val += 1;
carryover = false;
}
if (l1 != null)
{
l3.val += l1.val;
l1 = l1.next;
}
if (l2 != null)
{
l3.val += l2.val;
l2 = l2.next;
}
if (l3.val > 9)
{
l3.val -= 10;
carryover = true;
}
System.out.println(l3.val);
//create next 'link' in list
l3 = l3.next;
}
return head;
}
}
最佳答案
l3 = l3.next;
没有做你认为正在做的事情。l3.next
是null
,因此您将null
分配给l3
。null
在内存中不是一个特殊的位置,它只是指l3.next
,这意味着它没有指向任何东西。
所以在下一个循环中,当您执行null
时,您只是创建一个断开连接的节点。
您应该首先确保l3 = new ListNode(0);
指向一个节点,然后才能使用它。
所以,试试这个:
boolean first = true;
//if lists l1, l2 still have a value, append to l3
while (l1 != null || l2 != null)
{
// create the next node
if (!first) {
// create the next node and attach it to the current node
l3.next = new ListNode(0);
// we now work with the next node
l3 = l3.next;
} else {
first = false;
}
//if l1.val + l2.val >= 10 from last iteration, carry over 1
if (carryover)
{
l3.val += 1;
carryover = false;
}
if (l1 != null)
{
l3.val += l1.val;
l1 = l1.next;
}
if (l2 != null)
{
l3.val += l2.val;
l2 = l2.next;
}
if (l3.val > 9)
{
l3.val -= 10;
carryover = true;
}
System.out.println(l3.val);
}