Swap Nodes in Pairs

Description:

Given a linked list, swap every two adjacent nodes and return its head.
For example, Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes
itself can be changed.

#include <iostream>#include <limits.h>using namespace std;class LNode{public:    int val;    LNode *next;    LNode(int x):val(x),next(nullptr) { }};class Solution{    public:        LNode *swapNodes(LNode *L)        {            if (L == nullptr || L->next == nullptr || L->next->next == nullptr)                return L;            LNode *pre = L;            LNode *cur = pre->next;            LNode *nex = cur->next;            L = nex;            while (1)            {                pre->next = nex;                cur->next = nex->next;                nex->next = cur;                if (cur->next == nullptr)                    break;                pre = cur;                cur = cur->next;                if (cur->next == nullptr)                    break;                nex =  cur->next;            }            return L;        }};int main(){    LNode *List = new LNode(INT_MIN);    //create head node    cout<<"Input the number: "<<endl;    int num;    cin>>num;    LNode *ptr = List;    cout<<"Input the node: "<<endl;     //尾插法创建并输出链表;    for (int i = 0; i < num; ++i)    {        int value;        cin>>value;        ptr->next = new LNode(value);        ptr = ptr->next;        cout<<ptr->val;        if (i < num-1)            cout<<"->";    }    cout<<endl;    Solution solution;    LNode *ret = solution.swapNodes(List);    LNode *del = nullptr;    while(ret != nullptr)       //输出并释放链表    {        del = ret;        cout<<ret->val;        if (ret->next != nullptr)            cout<<"->";        ret = ret->next;        delete del;    }    delete List;    return 0;}