for (int i = 0; i < list->length; i++) { cout << cur->data << " "; cur = cur->next; } cout << endl; }

这段代码看起来是遍历链表并输出每个节点的数据。但是在循环中，`i` 没有进行自增操作，这会导致无限循环。应该将 `i++` 添加到循环语句中，即 `for (int i = 0; i < list->length; i++)`。同时，还需要检查链表是否为空，以及是否存在 `cur` 指针。完整的代码可能类似于下面这样：

if (list == nullptr || list->head == nullptr) {
    cout << "List is empty." << endl;
} else {
    Node* cur = list->head;
    for (int i = 0; i < list->length; i++) {
        cout << cur->data << " ";
        cur = cur->next;
    }
    cout << endl;
}

相关问题

帮我修改避免查重 #include <iostream> #include <algorithm> using namespace std; int vis[9], a[9], record[9], n; void DFS(int cur) { if (cur == n) { for (int i = 0; i < n; i++) cout << record[i] << " "; cout << endl; return; } for (int i = 0; i < n; i++) { if (vis[i]) continue; vis[i] = true; record[cur] = a[i]; DFS(cur + 1); vis[i] = false; } } int main() { cin >> n; for (int i = 0; i < n; i++) cin >> a[i]; sort(a, a + n); DFS(0); }

这段代码的主要问题在于没有考虑到输入的数字可能存在重复，因此会输出重复的排列。为了避免重复，可以在DFS的过程中添加一个判断，即如果当前数字和上一个数字相同且上一个数字还没有被使用过，就跳过当前数字的使用。修改后的代码如下：

#include <iostream>
#include <algorithm>
using namespace std;

int vis[9], a[9], record[9], n;

void DFS(int cur) {
    if (cur == n) {
        for (int i = 0; i < n; i++)
            cout << record[i] << " ";
        cout << endl;
        return;
    }
    for (int i = 0; i < n; i++) {
        if (vis[i] || (i > 0 && a[i] == a[i-1] && !vis[i-1]))
            continue;
        vis[i] = true;
        record[cur] = a[i];
        DFS(cur + 1);
        vis[i] = false;
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> a[i];
    sort(a, a + n);
    DFS(0);
}

修改后的代码会输出不重复的排列。

按之前题目要求补全以下代码/*-------------------------------------------------- 编程要求 函数node* insertBefore(node* head, node* tmp)的功能是将节点tmp插入到其对“偶节”点前面，并返回新链表。 如果不存在对偶节点，则将tmp插入链表头部。所谓的“对偶”节点是指他们的data值的和为0。 说明：链表head中节点的data值各不相同。 ------------------------------------------------*/ #include <iostream> using namespace std; struct node { int data; node* next; }; node* insertBefore(node* head, node* tmp); void output(node* head); int main() { int n; cin >> n; node* head = NULL, * tmp; bool flag; for (int i = 0; i < n; i++) { tmp = new node; cin >> tmp->data; tmp->next = head; head = tmp; } //output(head); // insert a node after its dual number tmp = new node; cin >> tmp->data; head = insertBefore(head, tmp); output(head); } void output(node* head) { while (head) { cout << head->data; head = head->next; if (head) cout << "->"; } cout << endl; }

好的，以下是完整的代码实现：

#include <iostream>
using namespace std;

struct node {
    int data;
    node* next;
};

node* insertBefore(node* head, node* tmp);
void output(node* head);

int main() {
    int n;
    cin >> n;
    node* head = NULL, * tmp;
    bool flag;
    for (int i = 0; i < n; i++) {
        tmp = new node;
        cin >> tmp->data;
        tmp->next = head;
        head = tmp;
    }
    //output(head);

    // insert a node after its dual number
    tmp = new node;
    cin >> tmp->data;
    head = insertBefore(head, tmp);
    output(head);
}

void output(node* head) {
    while (head) {
        cout << head->data;
        head = head->next;
        if (head) cout << "->";
    }
    cout << endl;
}

node* insertBefore(node* head, node* tmp) {
    if (head == nullptr) {
        return tmp;
    }

    node* pre = head;
    node* cur = head->next;
    int sum = tmp->data + head->data;

    while (cur != nullptr && sum != 0) {
        sum += cur->data;
        pre = cur;
        cur = cur->next;
    }

    if (sum == 0) {
        tmp->next = cur;
        pre->next = tmp;
    } else {
        tmp->next = head;
        head = tmp;
    }

    return head;
}