第1章第1节练习题7 顺序表的归并

问题描述

将两个有序的顺序表合并成一个新的有序顺序表，由函数返回结果顺序表

算法思想

本题实际就是归并排序的一种特殊情况，因为两个顺序表皆有序，这样我们只需要不断的取下两个顺序表中表头元素较小的那个数，然后将其存入新的顺序表，最后看哪个表还有剩余，将剩下的部分直接添加到新的顺序表后面即可。

算法描述

SqList* Merge(SqList *L1, SqList *L2)
{
    SqList *L = (SqList*)malloc(sizeof(SqList));
int i = 0, j = 0, k = 0;
while(i<L1->length&&j<L2->length){  
if (L1->data[i] <= L2->data[j]){
            L->data[k++] = L1->data[i++];
        }
else{
            L->data[k++] = L2->data[j++];
        }
    }
while (i != L1->length){
        L->data[k++] = L1->data[i++];
    }
while (j != L2->length){
        L->data[k++] = L1->data[j++];
    }
    L->length = k;
return L;
}

具体代码见附件

---

附件

#include<stdio.h>
#include<stdlib.h>
#define MaxSize 100
typedef int ElemType;

typedef struct{
    ElemType data[MaxSize];
int length;
}SqList;

SqList* Merge(SqList *, SqList *);
void print(SqList*);

int main(int argc, char* argv[])
{
    SqList SL1, SL2;
    SL1.length = 10;
    SL2.length = 10;
for (int i = 0; i < SL1.length; i++){
        SL1.data[i] = 2 * i + 1;
    }
for (int i = 0; i < SL2.length; i++){
        SL2.data[i] = 2 * i;
     }
    print(&SL1);
    print(&SL2);
    SqList *SL;
    SL = Merge(&SL1, &SL2);
    print(SL);
return 0;
}

SqList* Merge(SqList *L1, SqList *L2)
{
    SqList *L = (SqList*)malloc(sizeof(SqList));
int i = 0, j = 0, k = 0;
while(i<L1->length&&j<L2->length){  
if (L1->data[i] <= L2->data[j]){
            L->data[k++] = L1->data[i++];
        }
else{
            L->data[k++] = L2->data[j++];
        }
    }
while (i != L1->length){
        L->data[k++] = L1->data[i++];
    }
while (j != L2->length){
        L->data[k++] = L1->data[j++];
    }
    L->length = k;
return L;
}

void print(SqList *L){
for (int i = 0; i < L->length; i++){
printf("%4d", L->data[i]);
    }
printf("\n");
}