题解 | #从单向链表中删除指定值的节点# 构造，删除，打印

#include <iostream>
using namespace std;

struct ListNode {
    ListNode* next;
    int val;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int value) : val(value), next(nullptr) {}
    ListNode(ListNode* next) : val(0), next(next) {}
    ListNode(ListNode* next, int value) : val(value), next(next) {}
};

int main() {
    int num;
    // 输入个数
    cin >> num;
    ListNode* head = nullptr;
    while (num--) {
        // 构造头节点
        if (!head) {
            int headVal;
            cin >> headVal;
            head = new ListNode(headVal);
            continue;
        }
        int first, second;
        cin >> first >> second;
        // second值和链表中相等，将first插入到该链表后
        auto* p = new ListNode(first);
        auto* q = head;
        while (q) {
            if (q->val == second) {
                auto* tmp = q->next;
                q->next = p;
                p->next = tmp;
                break;
            }
            q = q->next;
        }
    }
    int delVal;
    // 待删除节点值
    cin >> delVal;
    ListNode dummy;
    dummy.next = head;
    auto* pre = &dummy;
    while (head) {
        if (head->val == delVal) {
            pre->next = head->next;
            break;
        }
        head = head->next;
        pre = pre->next;
    }
    head = dummy.next;
    while (head) {
        cout << head->val << " ";
        head = head->next;
    }
    return 0;
}
// 64 位输出请用 printf("%lld")