head=l1=ListNode() l1.next=ListNode() l1=l1.next return head.next
时间: 2024-02-29 20:55:06 浏览: 15
这段代码可以理解为创建一个包含两个节点的链表,并返回链表的第二个节点。具体来说,ListNode()表示创建一个新的节点,并将该节点的指针赋值给l1和head,即将head和l1节点都指向新创建的节点,成为链表的第一个节点。接着,l1.next=ListNode()表示在链表的第一个节点后面再创建一个新的节点,即链表的第二个节点。然后,l1=l1.next表示将当前节点l1指向链表的第二个节点,即将当前节点指针向后移动一位。最后,return head.next表示返回链表的第二个节点,即返回head节点指向的下一个节点,也就是链表的第二个节点。需要注意的是,在这个过程中,head节点一直指向链表的第一个节点,l1节点则在移动过程中指向链表的不同节点。
相关问题
用c语言写class ListNode: def __init__(self, x): self.val = x self.next = Nonedef mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode: if not l1: return l2 if not l2: return l1 if l1.val <= l2.val: head = l1 l1 = l1.next else: head = l2 l2 = l2.next cur = head while l1 and l2: if l1.val <= l2.val: cur.next = l1 l1 = l1.next else: cur.next = l2 l2 = l2.next cur = cur.next if l1: cur.next = l1 else: cur.next = l2 return head# 测试l1 = ListNode(1)l1.next = ListNode(3)l1.next.next = ListNode(5)l2 = ListNode(2)l2.next = ListNode(4)l2.next.next = ListNode(6)head = mergeTwoLists(l1, l2)while head: print(head.val, end=' ') head = head.next
以下是使用C语言实现的代码:
```c
#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if (!l1) return l2;
if (!l2) return l1;
struct ListNode *head, *cur;
if (l1->val <= l2->val) {
head = l1;
l1 = l1->next;
} else {
head = l2;
l2 = l2->next;
}
cur = head;
while (l1 && l2) {
if (l1->val <= l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) {
cur->next = l1;
} else {
cur->next = l2;
}
return head;
}
int main() {
struct ListNode *l1, *l2, *head;
l1 = (struct ListNode *)malloc(sizeof(struct ListNode));
l1->val = 1;
l1->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l1->next->val = 3;
l1->next->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l1->next->next->val = 5;
l1->next->next->next = NULL;
l2 = (struct ListNode *)malloc(sizeof(struct ListNode));
l2->val = 2;
l2->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l2->next->val = 4;
l2->next->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l2->next->next->val = 6;
l2->next->next->next = NULL;
head = mergeTwoLists(l1, l2);
while (head) {
printf("%d ", head->val);
head = head->next;
}
printf("\n");
return 0;
}
```
输出结果为:1 2 3 4 5 6
/** * 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 addTwoNumbers(ListNode l1, ListNode l2) { ListNode head =null, tail =null; int carry =0; while(l1 != null && l2 != null){ //取出元素 //计算总和 //创建新节点 //把新节点链接在tail后 //更新tail //更新 l1 l2 int num1 = l1!=null?l1.val:0; int num2 = l2!=null?l2.val:0; int sum = num1+num2+carry; if(head == null){ head = tail =new ListNode(sum%10); }else{ tail.next =new ListNode(sum%10); } tail = tail.next; if(l1 != null) l1 = l1.next; if(l2 != null) l2 = l2.next; carry = sum/10; } if (carry>0){ tail.next = new ListNode(carry); } return head;}}请问这段代码问什么报错NULL POINT EXCEPTION
这段代码可能会出现空指针异常的原因是在最后判断进位是否大于0时,如果carry大于0,应该新建一个节点并将其链接在tail后,然后返回head。但是在代码中,当carry大于0时,只是将tail.next赋值为新建的节点,但是tail.next.next是为空的,因此在访问tail.next.next时会出现空指针异常。
为了解决这个问题,可以在tail.next赋值时,同时更新tail的值。另外,在最后返回head之前,需要判断head是否为空,如果为空,则说明l1和l2的初始值都为null,直接返回null即可。
下面是修改后的代码:
```
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = null, tail = null;
int carry = 0;
while (l1 != null || l2 != null) {
int num1 = l1 != null ? l1.val : 0;
int num2 = l2 != null ? l2.val : 0;
int sum = num1 + num2 + carry;
carry = sum / 10;
ListNode node = new ListNode(sum % 10);
if (head == null) {
head = tail = node;
} else {
tail.next = node;
tail = tail.next;
}
if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}
if (carry > 0) {
tail.next = new ListNode(carry);
tail = tail.next;
}
if (head == null) {
return null;
}
return head;
}
```