【剑指offer】合并两有序单链表

转载请注明出处：http://blog.csdn.net/ns_code/article/details/25739727

    九度OJ上AC，采用归并的思想递归实现。

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。
(hint: 请务必使用链表。)输入：输入可能包含多个测试样例，输入以EOF结束。
对于每个测试案例，输入的第一行为两个整数n和m(0<=n<=1000, 0<=m<=1000)：n代表将要输入的第一个链表的元素的个数,m代表将要输入的第二个链表的元素的个数。
下面一行包括n个数t(1<=t<=1000000)：代表链表一中的元素。接下来一行包含m个元素，s(1<=t<=1000000)。输出：对应每个测试案例，
若有结果，输出相应的链表。否则，输出NULL。样例输入：

5 2
1 3 5 7 9
2 4
0 0

样例输出：

1 2 3 4 5 7 9
NULL

    AC代码：
#include<stdio.h>
#include<stdlib.h>

typedef int ElemType;

typedef struct Node
{
ElemType data;
struct Node *next;
}Node,*pNode;

/*
合并两个升序链表，合并后的链表依然升序排列
*/
pNode MergeList(pNode pHead1,pNode pHead2)
{
if(pHead1 == NULL)
return pHead2;
if(pHead2 == NULL)
return pHead1;

pNode pMergeHead = NULL;
if(pHead1->data > pHead2->data)
{
pMergeHead = pHead2;
pMergeHead->next = MergeList(pHead1,pHead2->next);
}
else
{
pMergeHead = pHead1;
pMergeHead->next = MergeList(pHead2,pHead1->next);
}
return pMergeHead;
}

int main()
{
int n,m;
while(scanf("%d %d",&n,&m) != EOF)
{
pNode pHead1 = NULL;
if(n > 0)
{
int i,data;
scanf("%d",&data);
pHead1 =(pNode)malloc(sizeof(Node));
if(pHead1 == NULL)
exit(EXIT_FAILURE);
pHead1->data = data;
pHead1->next = NULL;

pNode pCur = pHead1;
for(i=0;i<n-1;i++)
{
scanf("%d",&data);
pNode pNew =(pNode)malloc(sizeof(Node));
if(pNew == NULL)
exit(EXIT_FAILURE);
pNew->data = data;
pNew->next = NULL;
pCur->next = pNew;
pCur = pCur->next;
}
}

pNode pHead2 = NULL;
if(m > 0)
{
int i,data;
scanf("%d",&data);
pHead2 =(pNode)malloc(sizeof(Node));
if(pHead2 == NULL)
exit(EXIT_FAILURE);
pHead2->data = data;
pHead2->next = NULL;

pNode pCur = pHead2;
for(i=0;i<m-1;i++)
{
scanf("%d",&data);
pNode pNew =(pNode)malloc(sizeof(Node));
if(pNew == NULL)
exit(EXIT_FAILURE);
pNew->data = data;
pNew->next = NULL;
pCur->next = pNew;
pCur = pCur->next;
}
}

pNode pMergeHead = MergeList(pHead1,pHead2);
if(pMergeHead == NULL)
printf("NULL\n");
else
{
pNode pCur = pMergeHead;
while(pCur != NULL)
{
//这里主要时要注意输出的格式
if(pCur->next == NULL)
printf("%d\n",pCur->data);
else
printf("%d ",pCur->data);
pCur = pCur->next;
}
}
}
return 0;
}