剑指Offer系列---(19)合并两个排序的链表

1.题目描述：

输入两个递增排序的链表，合并这两个链表并使新链表中的结点仍然是按照递增排序的。链表结点定义如下：
struct ListNode
{
   int       m_nKey;
   ListNode* m_pNext;
};
2.分析：
1）考虑代码鲁棒性的问题。a.第一个链表为空链表;b.第二个链表为空链表;c.两个链表都为空链表.
3.源代码：

//  Copyright (c) 2015年 skewrain. All rights reserved.#include <iostream>#include <stdio.h>#include <stack>using namespace std;struct ListNode{    int m_nValue;    ListNode *m_pNext;};ListNode* CreateLink(int a[],int k){    ListNode *Head = NULL,*q = NULL;    for (int i=0; i<k; i++)    {        ListNode *pNew = new ListNode();        pNew->m_nValue = a[i];        pNew->m_pNext = NULL;                if(Head==NULL)        {            Head = pNew;            q = pNew;        }        else        {            q->m_pNext = pNew;            q = pNew;        }    }    return Head;}//从头到尾打印列表void printLink(ListNode *pHead){    cout<<"链表内容为:";    ListNode *p = pHead;    while (p) {        cout<<p->m_nValue<<" ";        p=p->m_pNext;    }    cout<<endl;}ListNode* Merge(ListNode* pHead1,ListNode* pHead2){    if(pHead1 == NULL)        return pHead2;    if(pHead2 == NULL)        return pHead1;        ListNode* pMergeHead = NULL;    if(pHead1->m_nValue < pHead2->m_nValue)    {        pMergeHead = pHead1;        pMergeHead->m_pNext = Merge(pHead1->m_pNext,pHead2);    }    else    {        pMergeHead = pHead2;        pMergeHead->m_pNext = Merge(pHead1, pHead2->m_pNext);    }    return pMergeHead;}int main(int argc,char *argv[]){    int a[]={1,3,5,7};    ListNode *ptr1 = CreateLink(a,4);    cout<<"合并之前的a";    printLink(ptr1);    int b[]={2,4,6,8};    ListNode *ptr2 = CreateLink(b,4);    cout<<"合并之前的b";    printLink(ptr2);    ListNode *c = Merge(ptr1,ptr2);    cout<<"合并之后的c";    printLink(c);    return 0;}