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。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
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){if(pHead1==NULL)return pHead2;if(pHead2==NULL)return pHead1;ListNode* mergeHead=pHead1;if(pHead1->val<pHead2->val){mergeHead=pHead1;mergeHead->next=Merge(pHead1->next,pHead2);}else{mergeHead=pHead2;mergeHead->next=Merge(pHead1,pHead2->next);}return mergeHead;}
};int main()
{Solution s;int n;struct ListNode *mergeHead=NULL;struct ListNode *pHead1=NULL;struct ListNode *pHead2=NULL;struct ListNode *p=NULL,*p2=NULL,*x=NULL;scanf("%d",&n);pHead1=(struct ListNode*)malloc(sizeof(struct ListNode));p=(struct ListNode*)malloc(sizeof(struct ListNode));pHead2=(struct ListNode*)malloc(sizeof(struct ListNode));p2=(struct ListNode*)malloc(sizeof(struct ListNode));pHead1->next=p;for(int i=0;i<n;i++){scanf("%d",&p->val);p->next=(struct ListNode*)malloc(sizeof(struct ListNode));x=p;p=p->next;}x->next=NULL;p=pHead1->next;pHead2->next=p2;for(int i=0;i<n;i++){scanf("%d",&p2->val);p2->next=(struct ListNode*)malloc(sizeof(struct ListNode));x=p2;p2=p2->next;}x->next=NULL;p2=pHead2->next;mergeHead=s.Merge(p,p2);while(mergeHead!=NULL){printf("%d",mergeHead->val);mergeHead=mergeHead->next;}return 0;
}