【数据结构周周练】003顺序栈与链栈

这次只有一道题，采用了五种算法来实现，一个是普通方法，不采用任何结构，一个是用顺序表，一个是用链表，一个是用顺序栈，一个是用链栈。但是结构定义和算法实现方法不唯一，如果大家有其他的算法实现，在下面评论哦！

目录

习题一：判断序列是否合法

1、题目

2、普通方法实现

3、顺序表实现

4、链表实现

5、顺序栈实现

6、链栈实现

---

习题一：判断序列是否合法

1、题目

I表示入栈操作，O表示出栈操作，栈的初始状态与终态均为空，由I，O组成序列，可以进行栈的操作的序列称为合法序列，否则称为不合法序列。
    判断给定的字符串是否合法。

A: IIIIOOOO

B: IOIOIOIO

C: IIOOOIOI

D: IIIOOIOO

2、普通方法实现

代码

#include

char A[] = { 'I','I','I','I','O','O','O','O' };
char B[] = { 'I','O','I','O','I','O','I','O' };
char C[] = { 'I','I','O','O','O','I','O','I' };
char D[] = { 'I','I','I','O','O','I','O','O' };

int Legal(char * Arr) {
	int k = 0;
	int i = 0;
	while (i < 8 && k >= 0) {
		if (Arr[i] == 'I')
			k++;
		else
			k--;
		i++;
	}
	if(k<0)
	{
		return 0;
	}
	return 1;
}

void main() {

	if (Legal(A))
		std::cout << "A 序列合法" << std::endl;
	else
		std::cout << "A 序列不合法" << std::endl;

	if (Legal(B))
		std::cout << "B 序列合法" << std::endl;
	else
		std::cout << "B 序列不合法" << std::endl;

	if (Legal(C))
		std::cout << "C 序列合法" << std::endl;
	else
		std::cout << "C 序列不合法" << std::endl;

	if (Legal(D))
		std::cout << "D 序列合法" << std::endl;
	else
		std::cout << "D 序列不合法" << std::endl;
}

执行结果

3、顺序表实现

代码

#define MAXLISTSIZE 20
#define LISTINCREMENT 5

#include 
#include
using namespace std;

char A[] = { 'I','I','I','I','O','O','O','O' };
char B[] = { 'I','O','I','O','I','O','I','O' };
char C[] = { 'I','I','O','O','O','I','O','I' };
char D[] = { 'I','I','I','O','O','I','O','O' };

typedef struct {
	char *elem;
	int length;
	int listsize;
}SqList;

int InitSqList(SqList & S) {
	S.elem = (char *)malloc(MAXLISTSIZE * sizeof(SqList));
	if (!S.elem)
	{
		cout << "空间分配失败" << endl;
		return OVERFLOW;
	}
	S.length = 0;
	S.listsize = MAXLISTSIZE;
	return 1;
}

int JudgeSqList(SqList S, char *A) {
	int j = 0;
	int k = 0;
	for (int i = 0; i < 8; i++)
	{
		
		if (A[i] == 'I') {
			S.elem[j] = A[i];
			j++;
		}
		else if (A[i] == 'O' && S.elem[k] == 'I')
		{
			S.elem[k] = A[i];
			k++;
		}
		else
		{
			return 0;
		}
	}
	return 1;
}

void main() {
	SqList LA, LB, LC, LD;
	InitSqList(LA);
	InitSqList(LB);
	InitSqList(LC);
	InitSqList(LD);

	if (JudgeSqList(LA, A))
		cout << "A 序列合法" << endl;
	else
		cout << "A 序列不合法" << endl;

	if (JudgeSqList(LB, B))
		cout << "B 序列合法" << endl;
	else
		cout << "B 序列不合法" << endl;

	if (JudgeSqList(LC, C))
		cout << "C 序列合法" << endl;
	else
		cout << "C 序列不合法" << endl;

	if (JudgeSqList(LD, D))
		cout << "D 序列合法" << endl;
	else
		cout << "D 序列不合法" << endl;
}

执行结果

 

4、链表实现

代码

#include 
#include
using namespace std;

char A[] = { 'I','I','I','I','O','O','O','O' };
char B[] = { 'I','O','I','O','I','O','I','O' };
char C[] = { 'I','I','O','O','O','I','O','I' };
char D[] = { 'I','I','I','O','O','I','O','O' };

typedef struct LNode {
	char elem;
	LNode * next;
}LNode,*LinkList;

int InitList(LinkList &L) {
	L = (LinkList)malloc(sizeof(LNode));
	if (!L)
	{
		cout << "空间分配失败" << endl;
		return OVERFLOW;
	}
	L->next = NULL;
}

int JudgeList(LinkList &L,char *A) {
	LinkList p = L;
	LinkList q = (LinkList)malloc(sizeof(LNode));
	q->elem = 'I';
	q->next = NULL;
	p->next = q;
	p = q;
	q = p->next;
	
	for (int i = 0; i < 8;i++)
	{
		if (!L->next)
			return 0;
		else if (A[i] == 'I')
		{
			q = (LinkList)malloc(sizeof(LNode));
			if (!q)
			{
				cout << "空间分配失败" << endl;
				return OVERFLOW;
			}
			q->elem = 'I';
			q->next = NULL;
			p->next = q;
			p = q;
		}
		else 
		{
			LinkList l = L->next;
			L->next = l->next;
			free(l);
		}
		
	}
	return 1;
}

void main() {
	LinkList LA, LB, LC, LD;
	InitList(LA);
	InitList(LB);
	InitList(LC);
	InitList(LD);

	if (JudgeList(LA,A))
		cout << "A 序列合法" << endl;
	else 
		cout << "A 序列不合法" << endl;

	if (JudgeList(LB, B))
		cout << "B 序列合法" << endl;
	else
		cout << "B 序列不合法" << endl;

	if (JudgeList(LC, C))
		cout << "C 序列合法" << endl;
	else
		cout << "C 序列不合法" << endl;

	if (JudgeList(LD, D))
		cout << "D 序列合法" << endl;
	else
		cout << "D 序列不合法" << endl;
}

执行结果

5、顺序栈实现

代码

#define MAXSTACKSIZE 20
#define STACKCREMENT 5
#include
#include

using namespace std;

typedef struct {
	int *base;
	int *top;
	int stacksize;

}SqStack;

int InitSqStack(SqStack &S) {
	S.base = (int*)malloc(MAXSTACKSIZE * sizeof(int));
	if (!S.base)
	{
		exit(OVERFLOW);
	}
	S.top = S.base;
	S.stacksize = MAXSTACKSIZE;
	return 1;
}

int Push(SqStack &S, int e) {
	if (S.top - S.base == S.stacksize) {
		S.base = (int *)realloc(S.base, (S.stacksize + STACKCREMENT) * sizeof(SqStack));
		if (!S.base)
		{
			cout << "空间分配失败" << endl;
			exit(OVERFLOW);
		}
		S.stacksize += STACKCREMENT;
	}
	*S.top++ = e;

	return 1;
}

int Pop(SqStack &S, int &e) {
	if (S.top == S.base)
		return 0;
	e = *--S.top;
	return 1;
}

int IsEmpty(SqStack S) {
	if (S.top==S.base)
	{
		return 1;
	}
	return 0;

}
int JudgeStack(SqStack &S, char *Arr,int e) {
	for (int i = 0; i < 8; i++)
	{
		if (Arr[i] == 'I')
			Push(S, e);
		else if (Arr[i] == 'O' && !IsEmpty(S))
			Pop(S, e);
		else
			return 0;
	}
	return 1;

}
void main() {
	char A[] = { 'I','I','I','I','O','O','O','O' };
	char B[] = { 'I','O','I','O','I','O','I','O' };
	char C[] = { 'I','I','O','O','O','I','O','I' };
	char D[] = { 'I','I','I','O','O','I','O','O' };
	SqStack SA, SB, SC, SD;
	int e = 1;
	InitSqStack(SA);
	InitSqStack(SB);
	InitSqStack(SC);
	InitSqStack(SD);
	
	if(JudgeStack(SA, A, e))
		cout<<"A数组合法。"<

执行结果

6、链栈实现

代码

#include
#include

using namespace std;

char A[] = { 'I','I','I','I','O','O','O','O' };
char B[] = { 'I','O','I','O','I','O','I','O' };
char C[] = { 'I','I','O','O','O','I','O','I' };
char D[] = { 'I','I','I','O','O','I','O','O' };
typedef struct LNode {
	int data;
	struct LNode *next;
}LNode,*LinkStack;

int InitStack(LinkStack &LS) {
	LS = (LinkStack)malloc(sizeof(LNode));
	if (!LS)
	{
		cout << "空间分配失败" << endl;
		return OVERFLOW;
	}
	LS->next = NULL;
	return 1;
}

int Push(LinkStack &LS, int e) {
	//LinkStack sp = LS;
	LinkStack sp = (LinkStack)malloc(sizeof(LNode));
	sp->data = e;
	sp->next = LS;
	LS = sp;
	return 1;
}

int Pop(LinkStack &LS, int &e) {
	LinkStack sp;
	if (LS->next)
	{
		sp = LS;
		LS = LS->next;
		free(sp);
		return 1;
	}
	return 0;
}

int JudgeStack(LinkStack &LS, char*Arr, int &e) {
	for (int i = 0; i < 8; i++)
	{
		if (Arr[i] == 'I')
			Push(LS, e);
		else if (Arr[i] == 'O' && LS->next)
			Pop(LS, e);
		else
			return 0;
	}
	return 1;
}

void main() {
	LinkStack LA, LB, LC, LD;
	InitStack(LA);
	InitStack(LB);
	InitStack(LC);
	InitStack(LD);
	int e = 1;

	if (JudgeStack(LA, A, e))
		cout << "A数组合法。" << endl;
	else
		cout << "A数组不合法。" << endl;

	if (JudgeStack(LB, B, e))
		cout << "B数组合法。" << endl;
	else
		cout << "B数组不合法。" << endl;

	if (JudgeStack(LC, C, e))
		cout << "C数组合法。" << endl;
	else
		cout << "C数组不合法。" << endl;

	if (JudgeStack(LD, D, e))
		cout << "D数组合法。" << endl;
	else
		cout << "D数组不合法。" << endl;
}

执行结果