Basic Calculator II - LeetCode 227

题目描述：

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:
"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5
Note: Do not use the eval built-in library function.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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分析：看到这道题，第一感觉是怎么又是计算器，都快写烂了。好吧，思路参照224 Basic Calculator

，唯一的差别就是本题少了括号，多了*和/，这就要求在做运算前必须先判断优先级。不多说，代码类似于今年华为的实习生笔试上机题——火星计算器，唯一的差别是运算符改变了下，用栈实现的流程是一模一样的。

/**/60ms//*/
class Solution {
public:
   bool isOperator(char ch){ //判断是否是运算符
		return (ch == '+' || ch == '-' || ch == '/' || ch == '*'||ch == '=');
    }

	char precede(char th1,char th2) /*判断两个运算符的优先级，>表示th1的优先级不低于th2的优先级*/
	{
		if((th1 == '*' || th1 =='/' )|| (th2 == '=') ||((th1 == '+' || th1 =='-') && (th2 == '+' || th2 =='-')))
			return '>';
		else
			return '<';
	}
	
	int operate(int a,char opd,int b)  /*进行 a op b 的运算*/
	{
		if(opd == '+')
			return a + b;
	    else if(opd == '-')
			return a - b;		
		else if(opd == '/' && b != 0)
			return a / b;	
		else if(opd == '*')
			return a * b;	
		else
			return -1;
	}

	void getTwo(stack &num,int &left,int &right) /*从num中获取两个操作数，注意左右和出栈顺序*/
	{
		right = num.top(); /*先出栈的为右操作数*/
		num.pop();
		left = num.top();
		num.pop();
	}
	int calculate(string s)  /*计算运算表达式s*/
	{
		s.push_back('=');
		int re = 0;
		stack operater;
		stack num;
		operater.push('=');
		char optrTop;
		for(string::size_type i = 0;i != s.size();)
		{
			if(s[i] == ' ')   /*跳过空格*/
			{
				i++;
				continue;
			}
			optrTop = operater.top();
			if(s[i] == '=' && optrTop == '=')  /*运算结束条件*/
			{
				//cout <<"optTop: "<< optrTop <')
					{
						int left ,right;
						getTwo(num,left,right);
						char theta = operater.top();
						operater.pop();
						//	cout <<"operater pop :" <