逆波兰表达式求解 和 括号匹配问题

#pragma once
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
#include<stdlib.h>
#define MAX 100
typedef char DataType;
typedef struct Stack
{
	int top;
	DataType stack[MAX];
}Stack;

void StackInit(Stack* s);
void StackPush(Stack* s, DataType data);
void StackPop(Stack* s);
DataType GetStackTop(Stack*s);
int GetStackSize(Stack*s);
int StackEmpty(Stack *s);
#include"Stack.h"
void StackInit(Stack* s)
{
	s->top = 0;
}
void StackPush(Stack* s, DataType data)
{
	if (s->top == MAX)
	{
		printf("栈已满!无法入栈元素\n");
		return;
	}
	
	(s->top)++;
	s->stack[s->top] = data;
}
void StackPop(Stack *s)
{
	if (s->top == 0)
	{
		printf("栈为空,无法出栈元素\n");
		return;
	}
	s->stack[s->top] = 0;
	(s->top)--;
}
DataType GetStackTop(Stack*s)
{
	if (s->top == 0)
	{
		printf("栈为空,无法获取栈顶\n");
		return 0;
	}
	return s->stack[s->top ];
}
int GetStackSize(Stack*s)
{
	return s->top;
}
int StackEmpty(Stack *s)
{
	if (s->top == 0)
	{
		return 1;
	}
	else
	{
		return 0;
	}
}

#逆波兰表达式

#include "Stack.h"
enum { Data, Add, Sub, Mul, Div }operate; //枚举

typedef struct Cell
{
	enum operate op;
	int data;
}cell;

int CalRPN(cell* RPN, int sz)//逆波兰表达式(后缀表达式)
{
	int i = 0;
	Stack s;
	assert(RPN);
	StackInit(&s);
	for (; i < sz; i++) 
	{
		int left = 0;
		int right = 0;
		switch (RPN[i].op)
		{
		case Data:
			StackPush(&s, RPN[i].data);      
			break;
		case Add:                           
			right = GetStackTop(&s);   
			StackPop(&s);
			left = GetStackTop(&s);
			StackPop(&s);
			StackPush(&s, right + left);      
			break;
		case Sub:                             
			right = GetStackTop(&s);
			StackPop(&s);
			left = GetStackTop(&s);
			StackPop(&s);
			StackPush(&s, left - right);
			break;
		case Mul:                           
			right = GetStackTop(&s);
			StackPop(&s);
			left = GetStackTop(&s);
			StackPop(&s);
			StackPush(&s, left * right);
			break;
		case Div:                            
			right = GetStackTop(&s);
			StackPop(&s);
			left = GetStackTop(&s);
			StackPop(&s);
			if (0 == right)
				printf("除数非法\n");
			else
				StackPush(&s, left / right);
			break;
		}
	}
	return GetStackTop(&s);
}
void RPNTest()
{   
	cell RPN[] = { { Data,12 },{ Data,3 },{ Data,4 },{ Add,0 },{ Mul,0 },{ Data,6 },
	{ Sub,0 },{ Data,8 },{ Data,2 },{ Div,0 },{ Add,0 } };
	int sz = sizeof(RPN) / sizeof(RPN[0]);
	int ret = CalRPN(RPN, sz);
	printf("%d\n", ret);
}
int main()
{

	RPNTest();
	system("pause");
	return 0;
}

#括号匹配

#include "Stack.h"
int  MatchBrackets(const char *str)
{
	int i = 0;
	Stack s;
	StackInit(&s);
	StackPush(&s, str[i]);
	int length = strlen(str);
	i++;
	while ( i<  length)
	{
		if (str[i] == '{' || str[i] == '(' || str[i] == '['|| str[i] == '}' || str[i] == ')' || str[i] == ']')
		{
			if (str[i] == '{' || str[i] == '(' || str[i] == '[')
			{
				StackPush(&s, str[i]);
			}
			else
			{
				if (!StackEmpty(&s))
				{
					return 0;
				}
				if (str[i] == '}' && GetStackTop(&s) == '{')
				{
					StackPop(&s);
				}
			else if (str[i] == ']' && GetStackTop(&s) == '[')
				{
					StackPop(&s);
				}
			else if (str[i] == ')' && GetStackTop(&s) == '(')
				{
					StackPop(&s);
				}
				else
				{
					StackPush(&s ,str[i]);
					break;
				}
			}
		}
			i++;	
	}
	if (StackEmpty(&s))
	{
		return 0;
	}
	return 1;
}
int main()
{
	char a[] = "(())abc{[(])}"; // 左右括号次序匹配
	char b[] = "(()))abc{[]}"; // 右括号多于左括号
	char c[] = "(()()abc{[]}"; // 左括号多于右括号
	char d[] = "(())abc{[]()}";
	int Flag = MatchBrackets(a);
	if (Flag == 1)
	{
		printf("括号匹配!\n");
	}
	else
	{
		printf("括号不匹配!\n");
	}
	 Flag = MatchBrackets(b);
	if (Flag == 1)
	{
		printf("括号匹配!\n");
	}
	else
	{
		printf("括号不匹配!\n");
	}
	 Flag = MatchBrackets(c);
	if (Flag == 1)
	{
		printf("括号匹配!\n");
	}
	else
	{
		printf("括号不匹配!\n");
	}
	 Flag = MatchBrackets(d);
	if (Flag == 1)
	{
		printf("括号匹配!\n");
	}
	else
	{
		printf("括号不匹配!\n");
	}
	system("pause");
	return 0;
}
    原文作者:括号匹配问题
    原文地址: https://blog.csdn.net/H_Strong/article/details/82225303
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