数据结构与算法编程题2

逆置线性表，使空间复杂度为 O(1)

#include 
using namespace std;

typedef int ElemType;
#define Maxsize 100
#define	OK 1
#define  ERROR 0
typedef struct SqList
{
	ElemType data[Maxsize];
	int length;
}SqList;

void Init_SqList(SqList& L)
{
	L.length = 0;
}

//逆置线性表，使空间复杂度为 O(1)
void inverse(SqList& L)
{
	int temp = 0;
	for (int i = 0; i <= (0 + L.length - 1) / 2; i++)
	{
		temp = L.data[i];
		L.data[i] = L.data[L.length - 1 - i];
		L.data[L.length - 1 - i] = temp;
	}
}

void inverse1(SqList& L)
{
	for (int i = 0, j = L.length - 1; i < j; i++, j--)
	{
		int t = L.data[i];
		L.data[i] = L.data[j];
		L.data[j] = t;
	}
}
//0 1 2 3 4  奇数
//5 6 7 1 10

//0 1 2 3 4 5偶数
//5 6 7 1 10 11
int main()
{
	SqList L;
	Init_SqList(L);
	cout << "//---------顺序表所存数据为奇数---------//" << endl;
	L.data[0] = 5;
	L.data[1] = 6;
	L.data[2] = 7;
	L.data[3] = 1;
	L.data[4] = 10;
	L.length = 5;
	for (int i = 0; i < L.length; i++)
		cout << L.data[i] << "     ";
	cout << endl;
	//inverse(L);
	inverse1(L);
	for (int i = 0; i < L.length; i++)
		cout << L.data[i] << "     ";
	cout << endl;
	cout << "//---------顺序表所存数据为偶数---------//"<<endl;
	L.data[0] = 5;
	L.data[1] = 6;
	L.data[2] = 7;
	L.data[3] = 1;
	L.data[4] = 10;
	L.data[5] = 11;
	L.length = 6;
	for (int i = 0; i < L.length; i++)
		cout << L.data[i] << "     ";
	cout << endl;
	//inverse(L);
	inverse1(L);
	for (int i = 0; i < L.length; i++)
		cout << L.data[i] << "     ";
	cout << endl;
	return 0;
}
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