Leetcode227 Basic Calculator II

描述：实现一个计算器，字符串中只包含非负整数，+，-，*，/和空格

Example1：    Input: "3+2*2" Output: 7

Example2：     Input: " 3/2 " Output: 1

这道题相对来说还是比较简单的，不需要考虑括号及其他运算，但是我做的答案还是有一些问题，先贴一个个人觉得比较好的代码

class Solution {
public:
    int calculate(string s) {
        stack<int> res;
        stringstream ss('+' + s);          //在整个字符串前面增加一个“+”作为第一个数字的符号
        int n;
        char op;
        while(ss >> op >> n) {             //stringstream这种用法值得学习，同时也排除了空格的影响
            if(op == '+') {
                res.push(n);
            }
            if(op == '-')
                res.push(-n);
            if(op == '*') {
                int ans = res.top() *n;
                res.pop();
                res.push(ans);
            }
            if(op == '/') {
                int ans = res.top() /n;
                res.pop();
                res.push(ans);
            }    
        }
        int ans=0;
        while(res.size()!=0) {ans+=res.top(); res.pop();}   
        return ans;
    }
};

下面是我的代码，建了两个栈——符号栈和数字栈，通过比较不同符号之间的优先级大小确定顺序，debug很久   一些小细节很容易出错

class Solution {
public:
	int cal(int num1, int num2, char c)
	{
		int num = 0;
		switch(c)
		{
		case '+':  num = num1 + num2; break;
		case '-':  num = num1 - num2; break;
		case '*':  num = num1 * num2; break;
		case '/':  num = num1 / num2; break;
		default:
			break;
		}
		return num;
	}
 	int calculate(string s) {
		map<char, int> priority;
		stack<int> num;
		stack<char> opt;
		priority['+'] = 0;
		priority['-'] = 0;
		priority['*'] = 1;
		priority['/'] = 1;
		int size = s.size();
		int i = 0;
		while(i < size)
		{
			while(i < size && s[i] == ' ')
            {
                i++;
            }
			if(i < size && s[i] >= '0' && s[i] <= '9')
			{
                int number = 0;
                while(i < size && s[i] >= '0' && s[i] <= '9')
                {
                    number = number * 10 + (s[i] - '0');
				    i++;
                }
                num.push(number);
				
			}
			else if(i < size)
			{
				if(opt.empty() || priority[s[i]] > priority[opt.top()])
					opt.push(s[i]);
				else
				{
					while(!opt.empty() && priority[s[i]] <= priority[opt.top()])
					{
						int num2 = num.top();
						num.pop();
						int num1 = num.top();
						num.pop();
						char c = opt.top();
						opt.pop();
						int res = cal(num1, num2, c);
						num.push(res);
						
					}
                    opt.push(s[i]);
				}
				

				i++;  
			}
		}
		
		while(!opt.empty() && !num.empty())
		{
			int num2 = num.top();
			num.pop();
			int num1 = num.top();
			num.pop();
			char c = opt.top();
			opt.pop();
			int res = cal(num1, num2, c);
			num.push(res);
		}
		return num.top();
	}
};

再贴一种代码   思路和上述我的思路一样，但是代码风格更好看一点：

class Solution {
public:
    bool cmp(char op1, char op2){
        if((op1 == '+' || op1 == '-') && (op2 == '*' || op2 == '/')){
            return false;
        }else if(op1 == '#'){
            return false;
        }else if(op2 == '#'){
            return true;
        }else{
            return true;
        }
    }
    int cal(int a, int b, int op){
		switch (op){
		case '+': return a + b;
		case '-': return a - b;
		case '*': return a * b;
		case '/': return a / b;
		default: return 0;
		}
	}
    int calculate(string s) {
        int ans = 0;
        int index = 0;
        s += "#";
        int len = s.length();
        stack<int> num;
        stack<char> opt;
        opt.push('#');
        while(index < len){
            while(index < len && s[index] == ' '){
                ++index;
            }
            if(index < len && s[index] >= '0' && s[index] <= '9'){
                int tmp = 0;
                while(index < len && s[index] >= '0' && s[index] <= '9'){
                    tmp = tmp * 10 + s[index] - '0';
                    ++index;
                }
                num.push(tmp);
                //cout<<"push: "<<tmp<<endl;
            }else if(index < len){
                char c = s[index++];
                //cout << "top "<<opt.top()<<" c "<<c<<endl;
                while(!opt.empty() && cmp(opt.top(), c)){
                    int b = num.top();
					num.pop();
					int a = num.top();
					num.pop();
					num.push(cal(a, b, opt.top()));
					//cout<<"push calc: "<<cal(a, b, opt.top())<<endl;
					opt.pop();
                }
                //cout<<"push: "<<c<<endl;
                opt.push(c);
            }
        }
        return num.top();
    }
};