def addAtIndex(self,index,val): if index < 0 or index > self.size: return -1 if index == self.size: self.addAtTail(val) return current = self.head for i in range(index-1): current = current.next new_node = Listnode(val,current,current.next) current.next.prev = new_node current.next = new_node self.size += 1 AttributeError: 'NoneType' object has no attribute 'prev' current.next.prev = new_node 请修改
时间: 2024-02-14 12:10:44 浏览: 66
这段代码中可能存在两个问题:
1. 如果链表为空,即 self.head 为 None,那么在第 9 行的 for 循环中会报错。因此需要在第 3 行加上判断链表是否为空的条件。
2. 如果插入的位置是最后一个结点,即 index == self.size,那么在第 7 行调用 addAtTail 方法后,函数应该返回而不是继续执行。因此需要在第 8 行加上 return 语句。
修改后的代码如下:
```
def addAtIndex(self, index, val):
if index < 0 or index > self.size:
return -1
if index == self.size:
self.addAtTail(val)
return
if self.head is None:
self.addAtHead(val)
return
current = self.head
for i in range(index-1):
current = current.next
new_node = ListNode(val, current, current.next)
current.next.prev = new_node
current.next = new_node
self.size += 1
```
这样就可以避免空指针异常和插入最后一个结点时的错误。
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
对下面代码每一步含义进行注释 def insert(self, val): if not self.root: self.root = TreeNode(val) return cur = self.root while cur: if val < cur.val: if not cur.left: cur.left = TreeNode(val) return else: cur = cur.left else: if not cur.right: cur.right = TreeNode(val) return else: cur = cur.right
这段代码是二叉搜索树的插入操作代码,具体注释如下:
```
def insert(self, val): # 定义插入操作函数,val为要插入的节点值
if not self.root: # 如果二叉搜索树为空,则将要插入的节点作为根节点
self.root = TreeNode(val) # 创建根节点
return # 返回
cur = self.root # 如果二叉搜索树不为空,则从根节点开始查找
while cur: # 进入循环,直到找到合适的插入位置或者查找到叶子节点
if val < cur.val: # 如果要插入的值小于当前节点值
if not cur.left: # 如果当前节点的左子节点为空,则将要插入的节点作为当前节点的左子节点
cur.left = TreeNode(val) # 创建左子节点
return # 返回
else: # 如果当前节点的左子节点不为空,则继续向左查找
cur = cur.left # 将当前节点设置为左子节点,继续查找
else: # 如果要插入的值大于等于当前节点值
if not cur.right: # 如果当前节点的右子节点为空,则将要插入的节点作为当前节点的右子节点
cur.right = TreeNode(val) # 创建右子节点
return # 返回
else: # 如果当前节点的右子节点不为空,则继续向右查找
cur = cur.right # 将当前节点设置为右子节点,继续查找
```
总体来说,这段代码的作用是将一个节点插入到二叉搜索树中,并保证插入后仍然是二叉搜索树。
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