两路归并的数组与链表的实现方法

#include<iostream>
#include<assert.h>
using namespace std;
struct node
{
    int val;
    node * next;
    node(int v)
    {
        val=v;
        next=null;
    }
};

node * merge(node* list1 , node * list2)
{
    assert(list1!=null&&list2!=null);
    node * res;
    if(list1->val<=list2->val)
    {
        res=list1;
        list1=list1->next;
    }
    else
    {
        res=list2;
        list2=list2->next;
    }
    node * p = res;
    node *p1 =list1,*p2 =list2;

    while(p1!=null&&p2!=null)
    {
        if(p1->val<=p2->val)
        {
            p->next=p1;
            p=p->next;
            p1=p1->next;
        }
        else
        {
            p->next=p2;
            p=p->next;
            p2=p2->next;
        }
    }

    while(p1!=null)
    {
        p->next=p1;
        p=p->next;
        p1=p1->next;
    }
    while(p2!=null)
    {
        p->next=p2;
        p=p->next;
        p2=p2->next;
    }
    return res;
}

int * merge(int * arr1,int la, int * arr2,int lb)
{
    int i=0,j=0;
    int * arr = new int[la+lb];
    int t=0;
    while(i<la&&j<lb)
    {
        if(arr1[i]<=arr2[j])
        {
            arr[t++]=arr1[i];
            i++;
        }
        else
        {
            arr[t++]=arr2[j];
            j++;
        }
    }
    while(i<la)
    {
        arr[t++]=arr1[i];
        i++;
    }
    while(j<lb)
    {
        arr[t++]=arr2[j];
        j++;
    }
    return arr;
}

void setlinkdata(node * & list1 , node * & list2)
{
    node * node1 = new node(2);
    node * node2 = new node(3);
    node * node3 = new node(7);
    node * node4= new node(9);
    node1->next=node2;
    node2->next=node3;
    node3->next=node4;
    list1=node1;

    node * node5 = new node(1);
    node * node6 = new node(4);
    node * node7 = new node(6);
    node * node8 = new node(8);
    node5->next=node6;
    node6->next=node7;
    node7->next=node8;
    list2=node5;
}

int main()
{
    node * list1;
    node * list2;
    setlinkdata(list1,list2);
    int arr1[]={1,6,15,17,19};
    int arr2[]={2,4,6,8,10};
    int * arr = merge(arr1,5,arr2,5);
    node * ans = merge(list1,list2);
    //print result
    int length=10;
    for(int i=0;i<10;i++)
    {
        cout<<*arr<<endl;
        arr++;
    }
    while(ans!=null)
    {
        cout<<ans->val<<endl;
        ans=ans->next;
    }
    return 0;
}