合并两个排序的链表

来源：互联网发布：网络问诊编辑：程序博客网时间：2024/06/04 08:42
#include<iostream>using namespace std;struct MyNode{    int data;    MyNode *next;};struct MyList{    MyNode *head;};void CreatMyList(MyList *ml,int n){    if (NULL == ml )    {        return;    }    ml->head = new MyNode;    MyNode *temp = ml->head;    for (int i=n; i<n+10; i=i+2)    {        MyNode *mn = new MyNode;        mn->data = i;        mn->next = NULL;        temp->next = mn;        temp = temp->next;    }}void DisplayMyList( const MyList * ml){    if (NULL == ml||NULL == ml->head)    {        return;    }    MyNode *temp = ml->head->next;    while(NULL != temp)    {        cout<<temp->data<<" ";        temp = temp->next;    }    cout<<endl;}MyNode * ReMergeNode(MyNode *mn1, MyNode *mn2){    if (NULL == mn1)    {        return mn2;    }    if (NULL == mn2)    {        return mn1;    }    MyNode *temp;    if (mn1->data <= mn2->data)    {        temp = mn1;        temp->next = ReMergeNode(mn1->next,mn2);    }else    {        temp = mn2;        temp->next = ReMergeNode(mn1,mn2->next);    }    return temp;}MyList * ReMergeList(MyList *ml1, MyList *ml2){    if (NULL == ml1 || NULL == ml1->head || NULL == ml1->head->next)    {        return ml2;    }    if (NULL == ml2 || NULL == ml2->head || NULL == ml2->head->next)    {        return ml1;    }    if (NULL != ml1->head->next && NULL != ml2->head->next)    {        ml1->head->next = ReMergeNode(ml1->head->next,ml2->head->next);        return ml1;    }}void main(void){    MyList *ml1 = new MyList;    MyList *ml2 = new MyList;    CreatMyList(ml1,0);    CreatMyList(ml2,10);    //ml1 = NULL;    //ml2 = NULL;    DisplayMyList(ml1);    DisplayMyList(ml2);    ml1 = ReMergeList(ml1,ml2);    DisplayMyList(ml1);    ml1 = MerList(ml1,ml2);    DisplayMyList(ml1);    getchar();}
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