ListNode* cur = dummy;这行代码报错
时间: 2023-08-06 11:09:34 浏览: 55
很抱歉,我犯了一个错误。在不使用mergeTwoLists函数的代码中,将cur初始化为dummy节点是不正确的,因为这样做会使dummy节点被包含在结果链表中。正确的做法是将cur初始化为dummy的下一个节点,例如:
```c++
ListNode* dummy = new ListNode();
ListNode* cur1 = l1;
ListNode* cur2 = l2;
ListNode* cur = dummy;
while (cur1 && cur2) {
if (cur1->val <= cur2->val) {
cur->next = cur1;
cur1 = cur1->next;
} else {
cur->next = cur2;
cur2 = cur2->next;
}
cur = cur->next;
}
cur->next = cur1 ? cur1 : cur2;
ListNode* res = dummy->next;
```
请注意,我们在while循环之前创建了dummy节点,并将其next指针初始化为NULL。在while循环中,我们使用cur1和cur2指针来遍历两个链表,并将较小的节点插入到cur的后面。每次插入后,我们将cur向后移动一个节点,以便于下一次插入。
最后,我们将剩余的链表插入到cur的后面,返回dummy节点的下一个节点作为结果链表。
相关问题
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = new ListNode(0); ListNode* cur = dummy; while (l1 != nullptr && l2 != nullptr) { ListNode** pp = (l1->val < l2->val) ? &l1 : &l2; cur->next = *pp; cur = cur->next; *pp = (*pp)->next; } cur->next = (l1 == nullptr) ? l2 : l1; ListNode* ans = dummy->next; delete dummy; return ans; }
这段代码是合并两个有序链表的代码,可以用来将两个有序链表合并成一个有序链表。主要思路是通过比较两个链表的节点的大小,将较小的节点添加到结果链表中,直到其中一个链表为空。最后将另一个链表剩下的节点直接添加到结果链表的末尾即可。
具体解释可以看下面的注释:
```
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(0); // 创建一个虚拟头节点
ListNode* cur = dummy; // cur 指向当前节点,初始时指向虚拟头节点
while (l1 != nullptr && l2 != nullptr) { // 当两个链表都非空时
ListNode** pp = (l1->val < l2->val) ? &l1 : &l2; // 找出较小的节点
cur->next = *pp; // 将较小的节点添加到结果链表中
cur = cur->next; // 更新当前节点为新加入的节点
*pp = (*pp)->next; // 将原链表中被添加的节点指针后移
}
cur->next = (l1 == nullptr) ? l2 : l1; // 将另一个非空的链表直接添加到结果链表末尾
ListNode* ans = dummy->next; // 记录结果链表的头节点
delete dummy; // 释放虚拟头节点的内存
return ans; // 返回结果链表的头节点
}
```
#include <iostream> using namespace std; struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = new ListNode(0); ListNode* cur = dummy; 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 dummy->next; } int main() { ListNode* l1 = new ListNode(1); l1->next = new ListNode(3); l1->next->next = new ListNode(5); ListNode* l2 = new ListNode(2); l2->next = new ListNode(4); l2->next->next = new ListNode(6); ListNode* merged = mergeTwoLists(l1, l2); while (merged) { cout << merged->val << " "; merged = merged->next; } return 0; }将这段代码中的有序表改成可自己输入
好的,你可以参考以下代码,其中cin输入的数字需要自己输入,每输入一个数字按回车键确认,输入完毕后输入-1结束输入。
```cpp
#include <iostream>
using namespace std;
struct ListNode {
int val;
ListNode* next;
ListNode(int x) : val(x), next(NULL) {}
};
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(0);
ListNode* cur = dummy;
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 dummy->next;
}
int main() {
ListNode* l1 = new ListNode(0);
ListNode* l2 = new ListNode(0);
ListNode* cur1 = l1;
ListNode* cur2 = l2;
int num;
cout << "Please enter the numbers for the first linked list: " << endl;
while (cin >> num && num != -1) {
cur1->next = new ListNode(num);
cur1 = cur1->next;
}
cout << "Please enter the numbers for the second linked list: " << endl;
while (cin >> num && num != -1) {
cur2->next = new ListNode(num);
cur2 = cur2->next;
}
ListNode* merged = mergeTwoLists(l1->next, l2->next);
while (merged) {
cout << merged->val << " ";
merged = merged->next;
}
return 0;
}
```
希望能够帮到你!