expression 5+4*(7-15) or have parenthesis in any order // 波兰表示法-优快云博客

本文链接：https://blog.youkuaiyun.com/benbenab/article/details/8302849

Answers write code in java /c for expression 5+4*(7-15) or have parenthesis in any order .

idea: Use two stack, one for number, one for operator. Every time push a number or an operation into stack.
1) After pushing a number, check the top of the operator, if it is ‘*’ or '/', pop two number and do the multiple or divide operation. Then
push the result into number stack.

2) if the operation is ')', then do the operation till meeting the '('. Then push the result into number stack.

After traversing all of the character, I will do all of left add and minus operation. when I have just one number into the numberStack and I am done evaluating the exp , I will return the head of the number Stack.

void Multi (stack<float>& num, stack<char>& operation){
	float value_1 = num.top();
	num.pop();
	float value_2 = num.top();
	num.pop();
	operation.pop();
	num.push(value_1*value_2);
}

void Divi (stack<float>& num, stack<char>& operation){
	float value_1 = num.top();
	num.pop();
	float value_2 = num.top();
	num.pop();
	operation.pop();
	num.push(value_2/value_1);
}

void addOrMinus(stack<float>& num, stack<char>& operation){
	float value_1 = num.top();
	num.pop();
	float value_2 = num.top();
	num.pop();
	if (operation.top() == '+')
		num.push(value_1+value_2);
	else
		num.push(value_2-value_1);
	operation.pop();
}

float readnum(string data,int& index){
	float value = 0;
	bool numeration = false;
	int location = 0;
	while (index < data.size()){
		if (data[index] == '.')
			numeration = true;
		else if (data[index] <= '9' && data[index] >= '0'){
			if (numeration == false)
				value = value*10 + (data[index] - '0');
			else{
				location++;
				value += ((data[index]- '0')*pow(10.0,-location));
			}
		}
		else
			break;
		index++;
	}
	index--;
	return value;
}
double express(string data){
	int len = data.size();
	stack<float> num;
	stack<char> operation; 
	operation.push('%');
	for(int i = 0; i<len;){
		if (data[i] <= '9'&& data[i]>= '0'){
			float value = readnum(data, i);
			num.push(value);
			if (operation.top() == '*')
				Multi (num, operation);
			else if (operation.top() == '/')
				Divi (num, operation);
		}
		else if (data[i] == ')'){
			while (operation.top() != '('){
				addOrMinus(num,operation);
			}
			operation.pop();
			if (operation.top() == '*')
				Multi (num, operation);
			else if (operation.top() == '/')
				Divi (num, operation);
		}
		else
			operation.push(data[i]);
		i++;
	}

	while(num.size() > 1){
		addOrMinus(num,operation);
	}

	return num.top();
}

出现的错误：1. operator 这个关键是不能用的，这是保留关键字

2. 注意value_1和value_2的顺序

3. pow include “math.h”

4. 处理数字是不要忘了 data[i] - '0'

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其实这道题就是波兰表示法。波兰法介绍：http://baike.baidu.com/view/552648.htm

如何由中序遍历改为波兰法：

　　1、建立运算符栈stackOperator用于运算符的存储，压入'\0'。

　　2、预处理表达式，正、负号前加0(如果一个加号（减号）出现在最前面或左括号后面，则该加号（减号）为正负号) 。（上面的解法忘了考虑这一条）

　　3、顺序扫描表达式，如果当前字符是数字（优先级为0的符号），则直接输出该数字；如果当前字符为运算符或括号（优先级不为0的符号），则判断第4点。

　　4、若当前运算符为'('，直接入栈；

　　若为')'，出栈并顺序输出运算符直到遇到第一个'('，遇到的第一个'(' 出栈但不输出；