第1章第1节练习题7 顺序表的归并
最后更新于：2022-04-01 19:50:22
## 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(ilength&&jlength){
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
#include
#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(ilength&&jlength){
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");
}
~~~
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