Leetcode160相交链表（记录）C++

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已于 2023-06-20 09:12:48 修改

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文章标签： c++ 链表开发语言数据结构

于 2023-06-19 19:31:25 首次发布

本文链接：https://blog.youkuaiyun.com/m0_73496101/article/details/131293819

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该代码实现了一个C++程序，用于寻找两个链表的交点。它定义了链表节点结构体ListNode，提供了构造链表的模板函数constructListnode，以及找到链表交点的getIntersectionNode方法。在主函数中，创建并连接两个有交点的链表，然后打印它们的交点。

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

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

int printflistnode(ListNode *head);

template <class T>
ListNode *constructListnode(vector<T> vec);

class Solution
{
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
    {
        ListNode *anode = new ListNode();
        anode = headA;
        ListNode *bnode = new ListNode();
        bnode = headB;
        while (anode != nullptr)
        {
            bnode = headB;
            while (bnode != nullptr)
            {
                if (anode == bnode)
                {
                    ListNode *res = new ListNode();
                    res=anode;
                    ListNode *temp_anode = new ListNode();
                    temp_anode = anode->next;
                    ListNode *temp_bnode = new ListNode();
                    temp_bnode = bnode->next;
                    while (temp_anode != nullptr)
                    {
                        if (temp_anode == temp_bnode)
                        {
                            temp_anode=temp_anode->next;
                            temp_bnode=temp_bnode->next;
                            continue;
                        }
                        else
                            break;
                    }
                    if (temp_anode == nullptr)
                        return res;
                }
                bnode = bnode->next;
            }
            anode = anode->next;
        }
        return nullptr;
    }
};

int main()
{
    // 两条具有交点的链表
    vector<int> veca = {4, 1, 8, 4, 5};
    ListNode *heada = constructListnode(veca);
    vector<int> vecb = {5, 0, 1};
    ListNode *headb = constructListnode(vecb);
    ListNode *temp_headb = headb;
    ListNode *temp_heada = heada;
    while (temp_headb->val != 1)
        temp_headb = temp_headb->next;
    while (temp_heada->val != 8)
        temp_heada = temp_heada->next;
    temp_headb->next = temp_heada;

    // printf链表
    printflistnode(heada);
    cout << endl;
    printflistnode(headb);

    cout << endl;

    Solution a;
    ListNode *res=a.getIntersectionNode(heada, headb);

    printflistnode(res);

    delete[] heada;
    delete[] headb;

    return 0;
}

int printflistnode(ListNode *head)
{
    ListNode *next_node = head;
    while (next_node != nullptr)
    {
        cout << next_node->val;
        next_node = next_node->next;
    }
    return 0;
}

template <class T>
ListNode *constructListnode(vector<T> vec)
{
    ListNode *head = nullptr;
    if (vec.size() > 0)
    {
        head = new ListNode(vec[0]);
        ListNode *last_node = new ListNode();

        for (int i = 1; i < vec.size(); i++)
        {
            if (i == 1)
            {
                head->next = new ListNode(vec[i]);
                last_node = head->next;
            }
            else
            {
                last_node->next = new ListNode(vec[i]);
                last_node = last_node->next;
            }
        }
    }
    return head;
}