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p18 第7题 c语言代码实现王道数据结构课后代码题_哔哩哔哩_bilibili 

本题代码如下

int merge(struct sqlist* A, struct sqlist* B, struct sqlist* C)
{
	if (A->length + B->length > C->length)//大于顺序表的最大长度
		return false;
	int i = 0;
	int j = 0;
	int k = 0;
	while (i < A->length && j < B->length)//循环，两两比较，小者存入结果表中
	{
		if (A->data[i] <= B->data[j])
			C->data[k++] = A->data[i++];
		else
			C->data[k++] = B->data[j++];
	}
	while (i < A->length)//还剩一个没有比较完的顺序表
		C->data[k++] = A->data[i++];
	while (j < B->length)
		C->data[k++] = B->data[j++];
	C->length = k;
	return true;
}

完整测试代码

#include<stdio.h>
#define true 1
#define false 0
#define Max 10
struct sqlist
{
	int data[Max];
	int length;
};
int merge(struct sqlist* A, struct sqlist* B, struct sqlist* C)
{
	if (A->length + B->length > C->length)//大于顺序表的最大长度
		return false;
	int i = 0;
	int j = 0;
	int k = 0;
	while (i < A->length && j < B->length)//循环，两两比较，小者存入结果表中
	{
		if (A->data[i] <= B->data[j])
			C->data[k++] = A->data[i++];
		else
			C->data[k++] = B->data[j++];
	}
	while (i < A->length)//还剩一个没有比较完的顺序表
		C->data[k++] = A->data[i++];
	while (j < B->length)
		C->data[k++] = B->data[j++];
	C->length = k;
	return true;
}
int main()
{
	struct sqlist A = { 1,3,5,7 };
	struct sqlist B = { 2,4,6 };
	A.length = 4;
	B.length = 3;
	struct sqlist C;
	C.length = 7;
	printf("A顺序表为：");
	for (int i = 0; i < A.length; i++)
		printf("%d", A.data[i]);
	printf("\nB顺序表为");
	for (int j = 0; j < B.length; j++)
		printf("%d", B.data[j]);
	if (!merge(&A, &B, &C))
	{
		printf("错误");
	}
	else
	{
		printf("\n合并后的顺序表C为：");
		for (int i = 0; i < C.length; i++)
			printf("%d", C.data[i]);
	}
		

	return 0;
}