笔试题四：带头节点head两个链表合并，并且有序

将2个分别有序的链表合并成一个链表，并且合并后的链表依然有序。

方法：初始化一个头结点head，然后比较list1,和list2链表的第一个节点，选择比较小的连接到head上去，如此往复。

#include "stdafx.h"#include<iostream>#include<queue>using namespace std;typedef struct node{        int data;        struct node *next;}Node,*List;List createList(int N,int multi){List head = (List)malloc(sizeof(Node));head->data = 0;head->next=NULL;int count = 0;List p = head;while(count<N){count++;List s = (List)malloc(sizeof(Node));s->data = count*multi;s->next = NULL;p->next = s;p = s;}return head;}void traverse(List head){if(head == NULL){return;}List p = head->next;while(p){cout<<p->data<<" ";p = p->next;}cout<<endl;}List unionList(List list1,List list2){if(list1 == NULL||list1->next==NULL)return list2;if(list2 == NULL||list1->next==NULL)return list1;list1 = list1->next;list2 = list2->next;List head = (List)malloc(sizeof(Node));List p = NULL;if(list1!=NULL && list2!=NULL)//初始化第一个节点{if(list1->data<=list2->data){head->next = list1;list1 = list1->next;}else{head->next = list2;list2 = list2->next;}}p = head->next;while(list1!=NULL && list2!=NULL)//主体部分{if(list1->data<=list2->data){p->next = list1;list1 = list1->next;p = p->next;}else{p->next = list2;list2 = list2->next;p = p->next;}}while(list1 != NULL)//如果list2已经完了,list1还有剩余{p->next = list1;list1 = list1->next;p = p->next;}while(list2 != NULL)//如果list1已经完了,list2还有剩余{p->next = list2;list2 = list2->next;p = p->next;}return head;}int main(){int N = 10;List head1 = createList(N,1);List head2 = createList(N,2);traverse(head1);traverse(head2);List head = unionList(head1,head2);traverse(head);    getchar();    return 0;}