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括号匹配 (链栈)

fbd4ffd0717b 2022-04-14 阅读 50
数据结构

描述

假设表达式中只包含三种括号:圆括号、方括号和花括号,它们可相互嵌套,如([{}])或({[][()]})等均为正确的格式,而{[]})}或{[()]或([]}均为不正确的格式.

输入一串括号
如果输入的右括号多余,输出:Extra right brackets
如果输入的左括号多余, 输出:Extra left brackets
如果输入的括号不匹配,输出:Brackets not match
如果输入的括号匹配,输出:Brackets match

输入

{{{{)))

输出

Brackets not match

样例输入

{([)]}

样例输出

Brackets not match

代码实现:

#include <iostream>
using namespace std;
typedef char ElemType;
typedef  struct LNode 
{
	ElemType data;
	struct LNode *next;
}LNode, *LinkStack;
void init(LinkStack &s);
bool empty(LinkStack s);
void push(LinkStack &s, ElemType e);
void display(LinkStack s);
void pop(LinkStack &s);
ElemType top(LinkStack s);
int size(LinkStack s);
void destroy(LinkStack &s);
int match(char *str);
int main(void)
{
	char str[100];
	
	cin >> str;
	int res = match(str);
	switch(res)
	{
		case 0: cout << "Brackets not match" << endl; break;
		case 1: cout << "Brackets match" << endl; break;
		case 2: cout << "Extra right brackets" << endl; break;
		case 3: cout << "Extra left brackets" << endl; break;
	}
	
	return 0;
}
void init(LinkStack &s)
{
	s = NULL;
}
bool empty(LinkStack s)
{
	return s == NULL;
}
void push(LinkStack &s, ElemType e)
{
	LNode *p = new LNode;
	p -> data = e;
	p -> next = s;
	s = p;
}
void display(LinkStack s)
{
	for(LNode *p = s; p; p = p -> next)
		cout << p -> data << " ";
	cout << endl;
}
void pop(LinkStack &s)
{
	if(empty(s)) return;
	LNode *p = s;
	s = s -> next;
	delete p;
}
ElemType top(LinkStack s)
{
	return s -> data;
}
int size(LinkStack s)
{
	int n = 0;
	LNode *p = s;
	while(p)
	{
		n++; p = p -> next;
	}
	return n;
}
void destroy(LinkStack &s)
{
	LNode *p = s;
	while(p)
	{
		LNode *tmp = p;
		p = p -> next;
		delete tmp;
	}
}
int match(char *str)
{
	LinkStack s;
	
	init(s);
	for(int i = 0; str[i]; i++)
	{
		if(str[i] == '(' || str[i] == '[' || str[i] == '{')
			push(s, str[i]);
		else if(str[i] == ')')
		{
			if(empty(s)) return 2;
			else if(top(s) != '(') return 0;
			else pop(s);
		}
		else if(str[i] == ']')
		{
			if(empty(s)) return 2;
			else if(top(s) != '[') return 0;
			else pop(s);
		}
		else if(str[i] == '}')
		{
			if(empty(s)) return 2;
			else if(top(s) != '{') return 0;
			else pop(s);
		}
	}
	if(empty(s)) return 1;
	return 3;
}
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