用栈实现把前缀变后缀和中缀的原代码

[gudaodeai]

给你的是中缀表达式，怎样变成前缀呢？前缀又怎么变成后缀呢，难死我了。急急急。

[火龙]

给你一个转后缀的，你自己去写另一个

[火龙]

栈声明：
文件名是：Stack.h

#ifndef STACK_H
#define STACK_H

template <class T>
class Stack
{
public:
	Stack (int = 20);
	~Stack () {delete [] stackPtr;}
	void push (const T&);
	void pop (T&);
	int isEmpty () const {return top == -1;}
	int isFull () const {return top == size - 1;}
	void addTop (int);
	void detructTop (int);
	int getTop () {return top;}
	T getStack () {return stackPtr[top];}
private:
	int size;
	int top;
	T *stackPtr;
};

template <class T>
Stack<T>::Stack (int s)
{
	size = s;
	top = -1;
	stackPtr = new T [size];
}

template <class T>
void Stack<T>::push (const T &item)
{
	if (!isFull())
	{
		stackPtr[++top] = item;
	}
	else
		cout << "ERRO!" << endl;
}

template <class T>
void Stack<T>::pop (T &item)
{
	if (!isEmpty ())
	{
		item = stackPtr [top--];
	}
}

template <class T>
void Stack<T>::addTop (int s)
{
	top = top + s;
}

template <class T>
void Stack<T>::detructTop (int s)
{
	top = top - s;
}

#endif

[火龙]

主函数所在文件：

#include <iostream>
#include "Stack.h"
using namespace std;

int func (char c)   //返回栈顶运算符的优先级
{
	int top = 0;
	switch (c)
	{
		case '+':
		case '-':
			top = 1;
			break;
		case '*':
		case '/':
			top = 2;
			break;
		case ')':
			top = 3;
			break;
		case '(':
			top = 0;
			break;
		default:
			break;
	}

	return top;

}

int main ()
{
	cout << "请输入中缀表达式：" << endl;
	char *string;
	Stack<char> str1(1000), str2(1000);
	string = new char [1000];
	cin >> string;
	int i = 0;
	int *order;
	order = new int [1000];


	while (string [i] != '\0')  //判定输入表达式内运算符的优先级
	{
		if (string [i] == '+' || string [i] == '-')
			order [i] = 1;
		else if (string [i] == '*' || string [i] == '/')
			order [i] = 2;
		else if (string [i] == ')')
			order [i] = 3;
		else if (string [i] == '(')
			order [i] = 0;
		else
		{ }
		i ++;
	}

	i = 0;
	int top;
	char z;

	while (string [i] != '\0')
	{
		if (string [i] == '+' || string [i] == '-' 
			|| string [i] == '*' || string [i] == '/' 
			|| string [i] == '(' || string [i] == ')')
		{
			if (!str2.isEmpty ())
			{
				if (order [i] <= top)
				{
					while (order[i] <= top && !str2.isEmpty () && string[i] != '(')
					{
						str2.pop (z);
						str1.push (z);
					//	str2.push (string [i]);
						top = func (str2.getStack());
					}
					str2.push (string [i]);
				}
				else
				{
					str2.push (string [i]);
					top = func (str2.getStack ());
				}
			}
			else
			{
				str2.push (string [i]);
				top = func (str2.getStack());
			}
		}
		else 
		{
			str1.push (string [i]);
		}

		i ++;
	}


	cout << "\n输出后缀表达式：\n";

	while (!str1.isEmpty ())
	{
		str1.pop (z);
		str2.push (z);
	}

	while (!str2.isEmpty ())
	{
		str2.pop (z);
		if (z != '(' && z != ')')
			cout << z;
	}

	delete [] string;
	delete [] order;

	cout << endl;

	system ("pause");

	return 0;
}

[gudaodeai]

中缀变后缀，我也会，但前缀变后缀呢？我不会，你会不？？？？？

[gudaodeai]

记得要用栈实现

[火龙]

[Quote=引用 4 楼 gudaodeai 的回复:]

中缀变后缀，我也会，但前缀变后缀呢？我不会，你会不？？？？？
[/Quote]
既然都会前者了后者还不是顺水推舟的事情？你的输出只不过是把顺序调动了一下罢了，如果还是不会那只能说你还没弄懂这个原理，这就跟树的前序中序和后序遍历一样的过程

[gudaodeai]

好啊，我给你一个前缀。+*ab*-c/def。现在就这个式子，大虾，你能把它变成ab*cde/-f*+吗？记得用栈来实现哦。大虾，你会不？？？？？。

[尘缘udbwcso]

//表达式求值
#include <stdio.h>
#include <malloc.h>
#include <string.h>
/*
*功能:根据运算符计算
*参数:a, b参与运算的数, ch运算符
*返回值:计算结果,操作符错误则返回0
*/
int cal(int a, char ch, int b)
{
	switch(ch)
	{
	case '+':
		return a+b;
		break;
	case '-':
		return a-b;
		break;
	case '*':
		return a*b;
		break;
	case '/':
		return a/b;
		break;
	}
	return 0;
}
/*
*功能:计算表达式的值(用数组模拟栈)
*参数:表达式字符串
*返回值:计算结果
*/
int evaluateExpression(char *str)
{
	int i = 0, result, numSub = 0, operSub = 0;
	int tmp, len = strlen(str);
	int *operand = (int*)malloc(sizeof(int)*len);
	char *operat = (char*)malloc(sizeof(char)*len);
	while(str[i] != '\0')
	{
		switch(str[i])
		{
		case '+':
			while(operSub > 0 && operat[operSub-1] != '(')
			{
				printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);
				operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);
				printf("%d\n", operand[numSub-2]);
				--numSub;
				--operSub;
			}
			operat[operSub++] = '+';
			break;
		case '-':
			while(operSub > 0 && operat[operSub-1] != '(')
			{
				printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);
				operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);
				printf("%d\n", operand[numSub-2]);
				--numSub;
				--operSub;
			}
			operat[operSub++] = '-';
			break;
		case '*':
			if(str[i+1] >= '0' && str[i+1] <= '9')
			{
				tmp = 0;
				while(str[i+1] >= '0' && str[i+1] <= '9')
				{
					tmp = tmp * 10 + str[i+1] - '0';
					++i;
				}
				--i;
				printf("%d * %d = ", operand[numSub-1], tmp);
				operand[numSub-1] = cal(operand[numSub-1], '*', tmp);
				printf("%d\n", operand[numSub-1]);
				++i;
			}
			else
				operat[operSub++] = '*';
			break;
		case '/':
			if(str[i+1] >= '0' && str[i+1] <= '9')
			{
				tmp = 0;
				while(str[i+1] >= '0' && str[i+1] <= '9')
				{
					tmp = tmp * 10 + str[i+1] - '0';
					++i;
				}
				--i;
				printf("%d / %d = ", operand[numSub-1], tmp);
				operand[numSub-1] = cal(operand[numSub-1], '/', tmp);
				printf("%d\n", operand[numSub-1]);
				++i;
			}
			else
				operat[operSub++] = '/';
			break;
		case '(':
			operat[operSub++] = '(';
			break;
		case ')':
			while(operat[operSub-1] != '(')
			{
				printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);
				operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);
				printf("%d\n", operand[numSub-2]);
				--numSub;
				--operSub;
			}
			--operSub;
			break;
		default:
			tmp = 0;
			while(str[i] >= '0' && str[i] <= '9')
			{
				tmp = tmp * 10 + str[i] - '0';
				++i;
			}
			--i;
			operand[numSub++] = tmp;
			break;
		}
		++i;
	}
	while(numSub > 1 && operSub >= 1)
	{
		printf("%d %c %d = ", operand[numSub-2], operat[operSub-1], operand[numSub-1]);
		operand[numSub-2] = cal(operand[numSub-2], operat[operSub-1], operand[numSub-1]);
		printf("%d\n", operand[numSub-2]);
		--numSub;
		--operSub;
	}
	result = operand[numSub-1];
	free(operand);
	free(operat);
	return result;
}
int main()
{
	char *str = "225/15-20+(4-3)*2";
	int result;
	printf("计算过程:\n");
	result = evaluateExpression(str);
	printf("计算结果:result = %d\n", result);
	return 0;
}

计算过程:
225 / 15 = 15
15 - 20 = -5
4 - 3 = 1
1 * 2 = 2
-5 + 2 = -3
计算结果:result = -3

[gudaodeai]

大虾，给你出道题。输入前缀表达式：+*ab*-c/def.编个程序，让其输出是：ab*cde/-f*+（这是后缀表达式）。大虾会不？？？我急用

[火龙]

[Quote=引用 9 楼 gudaodeai 的回复:]

大虾，给你出道题。输入前缀表达式：+*ab*-c/def.编个程序，让其输出是：ab*cde/-f*+（这是后缀表达式）。大虾会不？？？我急用
[/Quote]

楼主，你这样做已经非常不厚道了，我给的和楼上楼上给的都是关于这个的源码，对于这样代码长度的CSDN论坛一般不会给源码的，而且CSDN以及各种论坛不欢迎找人写作业的，既然你要的其中一种情况已经给出的完整的源码你要是自己思考了一定马上可以举一反三，但是你一再的要求源码，根本不理会别人给思路。如果你坚持要求我们帮你写作业请自行离开，这里不是你找人帮忙写作业的地方。