16、剑指offer--合并两个排序的链表

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

 

解题思路：

1）定义一个头指针head，定义pNode为当前操作结点

2）判断链表1和链表2，将小的连接在pNode->next上，然后pNode = pNode->next

3）将剩下的非空链表链接上

4）返回head->next

#include <iostream>
#include <malloc.h>
using namespace std;
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode *head = new ListNode(-1);//头指针
        ListNode *pNode = head;//当前操作结点
        while(pHead1 != NULL && pHead2 != NULL)
        {
            if(pHead1->val <= pHead2->val)//小的链接在链表上
            {
                pNode->next = pHead1;
                pNode= pNode->next;
                pHead1 = pHead1->next;
            }
            else
            {
                pNode->next = pHead2;
                pNode = pNode->next;
                pHead2 = pHead2->next;
            }
        }
        //剩余的非空链表全部链接上
        while(pHead1 != NULL)
        {
            pNode->next = pHead1;
            pNode= pNode->next;
            pHead1 = pHead1->next;
        }
        while(pHead2 != NULL)
        {
            pNode->next = pHead2;
            pNode= pNode->next;
            pHead2 = pHead2->next;
        }
        return head->next;
    }
};
ListNode *CreateList(int n)
{
    ListNode *head;
    ListNode *p,*pre;
    int i;
    head=(ListNode *)malloc(sizeof(ListNode));
    head->next=NULL;
    pre=head;
    for(i=1;i<=n;i++)
    {
        p=(ListNode *)malloc(sizeof(ListNode));
        cin>>p->val;
        pre->next=p;
        pre=p;
    }
    p->next=NULL;

    return head->next;
}
/*-------------------------输出链表-----------------------------------*/
void PrintList(ListNode *h)
{
    ListNode *p;

    p=h;//不带空的头结点
    while(p)
    {
        cout<<p->val<<" ";
        p=p->next;
        cout<<endl;
    }
}
int main()
{
    int n1;
    int n2;
    ListNode *h1;
    ListNode *h2;
    ListNode *h;
    cout<<"输入链表1的结点数目"<<endl;
    cin>>n1;
    h1 = CreateList(n1);
    cout<<"链表1为："<<endl;
    PrintList(h1);
    cout<<"输入链表2的结点数目"<<endl;
    cin>>n2;
    h2 = CreateList(n2);
    cout<<"链表2为："<<endl;
    PrintList(h2);
    Solution s;
    h = s.Merge(h1,h2);
    cout<<"合并后链表为: "<<endl;
    PrintList(h);
    return 0;
}