ListNode *cur=h;
时间: 2024-03-15 11:43:23 浏览: 22
这段代码是将链表头节点赋值给 cur 变量,用于在遍历链表时跟踪当前结点。在链表操作中,通常需要使用一个指针变量来遍历链表,以便对每个结点进行具体的操作。由于链表是一种动态数据结构,其内存空间不是连续存储的,因此需要使用指针来访问每个结点,而 cur 变量就是用来存储当前指向的结点地址的指针变量。在这段代码中,cur 指针一开始指向哑结点 h,随着遍历链表的进行,cur 指针会指向新创建的节点,以便将其加入到结果链表中。
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#include <iostream> #include "listnode.h" using namespace std; struct listNode { listNode * prev, * next; int val; listNode(): val(0), prev(nullptr), next(nullptr){} listNode(int v, listNode *p, listNode *n): val(v), prev(p), next(n) { if (prev != nullptr) prev->next = this; if (next != nullptr) next->prev = this; } }; class OrderedList { protected: listNode * root; public: OrderedList(){root=new listNode();} ~OrderedList(){ listNode* cur=root; if(cur->prev!=nullptr)cur=cur->prev; if(cur!=nullptr){listNode*tmp=cur;cur=cur->next;delete tmp;}} virtual void insert(int val)=0; void printList() const { listNode* cur=root; if(cur->prev!=nullptr)cur=cur->prev; if(cur!=nullptr){cout<<cur->val<<' ';cur=cur->next;} } }; class AscendOrderedList:public OrderedList { public: AscendOrderedList(){root=new listNode();} ~AscendOrderedList(){ listNode* cur=root; if(cur->prev!=nullptr)cur=cur->prev; if(cur!=nullptr){listNode*tmp=cur;cur=cur->next;delete tmp;}} virtual void insert(int v){ listNode*pre=root; while(v>pre->val&&pre!=nullptr){pre=pre->next;} listNode*ins=new listNode(v, pre,pre->next); pre->next=ins; ins->next->prev=ins; } }; class DescendOrderedList:public OrderedList { public: DescendOrderedList(){root=new listNode();} ~DescendOrderedList(){ listNode* cur=root; if(cur->prev!=nullptr)cur=cur->prev; if(cur!=nullptr){listNode*tmp=cur;cur=cur->next;delete tmp;}} virtual void insert(int v){ listNode*pre=root; while(v<pre->val&&pre!=nullptr){pre=pre->next;} listNode*ins=new listNode(v, pre, pre->next); pre->next=ins; ins->next->prev=ins; } }为什么运行崩溃
这段代码中可能会有多个问题,具体原因需要根据您的输入、程序运行环境以及错误提示等来确定。以下是一些可能的问题和建议:
1. 在 `listNode` 的构造函数中,需要判断 `prev` 和 `next` 是否为空指针,才能为它们的 `prev` 和 `next` 成员变量赋值。修改代码如下:
```
listNode(int v, listNode* p, listNode* n) : val(v), prev(p), next(n) {
if (prev != nullptr) {
prev->next = this;
}
if (next != nullptr) {
next->prev = this;
}
}
```
2. 在 `OrderedList` 和其子类的析构函数中,需要循环删除链表中的所有节点。修改代码如下:
```
~OrderedList() {
listNode* cur = root;
while (cur->next != nullptr) {
cur = cur->next;
delete cur->prev;
}
delete cur;
}
~AscendOrderedList() {
listNode* cur = root;
while (cur->next != nullptr) {
cur = cur->next;
delete cur->prev;
}
delete cur;
}
~DescendOrderedList() {
listNode* cur = root;
while (cur->next != nullptr) {
cur = cur->next;
delete cur->prev;
}
delete cur;
}
```
3. 在 `AscendOrderedList` 和 `DescendOrderedList` 的 `insert()` 方法中,需要先判断当前节点是否为空指针,否则会出现访问空指针的错误。修改代码如下:
```
void insert(int v) {
listNode* pre = root;
while (pre->next != nullptr && v > pre->val) {
pre = pre->next;
}
listNode* ins = new listNode(v, pre, pre->next);
pre->next = ins;
if (ins->next != nullptr) {
ins->next->prev = ins;
}
}
```
```
void insert(int v) {
listNode* pre = root;
while (pre->next != nullptr && v < pre->val) {
pre = pre->next;
}
listNode* ins = new listNode(v, pre, pre->next);
pre->next = ins;
if (ins->next != nullptr) {
ins->next->prev = ins;
}
}
```
以上是可能的问题和建议,如果您还有其他问题或者需要更多帮助,请提供更多详细的信息和错误提示,以便我们更好地帮助您解决问题。
用C语言改写,将打印输出的结果中重复的数去掉#include <stdio.h>#include <stdlib.h>struct ListNode { int val; struct ListNode *next;};struct ListNode* mergeList(struct ListNode* l1, struct ListNode* l2){ struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode)); // 新建一个虚拟头节点 dummy->val = 0; dummy->next = NULL; struct 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; // 将剩余的节点加入到新链表中 if (l2) cur->next = l2; return dummy->next;}int main() { int n, m, num; scanf("%d%d", &n, &m); struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* p1 = l1; struct ListNode* p2 = l2; for (int i = 0; i < n; i++) { scanf("%d", &num); p1->next = (struct ListNode*)malloc(sizeof(struct ListNode)); p1->next->val = num; p1->next->next = NULL; p1 = p1->next; } for (int i = 0; i < m; i++) { scanf("%d", &num); p2->next = (struct ListNode*)malloc(sizeof(struct ListNode)); p2->next->val = num; p2->next->next = NULL; p2 = p2->next; } struct ListNode* res = mergeList(l1->next, l2->next); while (res) { printf("%d ", res->val); res = res->next; } return 0;}
以下是用 C 语言改写,并去掉重复数的代码:
#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode* mergeList(struct ListNode* l1, struct ListNode* l2){
struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode)); // 新建一个虚拟头节点
dummy->val = 0;
dummy->next = NULL;
struct 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; // 将剩余的节点加入到新链表中
if (l2) cur->next = l2;
return dummy->next;
}
int main() {
int n, m, num;
scanf("%d%d", &n, &m);
struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* p1 = l1;
struct ListNode* p2 = l2;
for (int i = 0; i < n; i++) {
scanf("%d", &num);
p1->next = (struct ListNode*)malloc(sizeof(struct ListNode));
p1->next->val = num;
p1->next->next = NULL;
p1 = p1->next;
}
for (int i = 0; i < m; i++) {
scanf("%d", &num);
p2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
p2->next->val = num;
p2->next->next = NULL;
p2 = p2->next;
}
struct ListNode* res = mergeList(l1->next, l2->next);
// 去除重复数
int hash[10001] = {0}; // 哈希表,假设节点值范围在 [0, 10000]
struct ListNode* cur = res;
struct ListNode* pre = NULL;
while (cur) {
if (hash[cur->val]) { // 如果当前节点值已经出现过,则删除当前节点
pre->next = cur->next;
free(cur);
cur = pre->next;
} else { // 如果当前节点值没有出现过,则将其加入哈希表,并继续遍历下一个节点
hash[cur->val] = 1;
pre = cur;
cur = cur->next;
}
}
// 输出结果
cur = res;
while (cur) {
printf("%d ", cur->val);
cur = cur->next;
}
// 释放内存
cur = res;
while (cur) {
struct ListNode* temp = cur;
cur = cur->next;
free(temp);
}
return 0;
}
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