題意：將倆個遞增連結串列合併成非遞減連結串列，注意提交程式碼時只需提交合並函式！

思路：就是最基本的連結串列操作。。。

程式碼：

#include <stdio.h>
#include <stdlib.h>

typedef int ElementType;
typedef struct Node* PtrToNode;
struct Node {
    ElementType Data;
    PtrToNode   Next;
};

typedef PtrToNode List;

List Read(); /* 細節在此不表 */
void Print( List L ); /* 細節在此不表；空連結串列將輸出NULL */

List Merge( List L1, List L2 );

int main()
{
    List L1, L2, L;
    L1 = Read();
    L2 = Read();
    L = Merge(L1, L2);
    Print(L);
    Print(L1);
    Print(L2);
    return 0;
}
List Read()
{
    int n,i,fig;
    List Head,r,s;
    Head = (List)malloc(sizeof(List));
    r = Head;
    //Head = r;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&fig);
        s = (List)malloc(sizeof(List));
        s->Data = fig;
        r->Next = s;
        r = s;
    }
    r->Next = NULL;
    return Head;
}
void Print( List L )
{
    List t = NULL;
    t = (List)malloc(sizeof(List));

    L = L->Next;
    if(L == NULL) {printf("NULL\n");return;}
    while(L != NULL){
        printf("%d ",L->Data);
        L = L->Next;
    }printf("\n");
    L = t;
}
List Merge( List L1, List L2 )
{
    List L,p,t,tL1,tL2;
    L = (List)malloc(sizeof(List));
    p = L;
    tL1 = L1->Next;tL2 = L2->Next;
    while(tL1 != NULL || tL2 != NULL){
        t = (List)malloc(sizeof(List));
        if(tL2 == NULL || tL1 != NULL && tL1->Data <= tL2->Data){
            t->Data = tL1->Data;
            p ->Next = t;
            p = t;
            tL1 = tL1->Next;
        }
        else{
            t->Data = tL2->Data;
            p->Next = t;
            p = t;
            tL2 = tL2->Next;
        }
    }
    p->Next = NULL;
    L1->Next = NULL;//將連結串列頭結點斷開
    L2->Next = NULL;
    return L;
}