Leetcode 224. Basic Calculator

224. Basic Calculator

• Total Accepted: 32728
• Total Submissions: 139195
• Difficulty: Hard

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

Note: Do not use the eval built-in library function.

思路：见代码。

代码：

num表示当前项的数值，res表示之前所有项的和，sign表示的前一项的符号。

class Solution {
public:
    int calculate(string s) {
        stack<int> nums;
        int sign=1,res=0,num=0;
        for(int i=0;i<s.size();i++){
            if(isdigit(s[i])){
                num*=10;
                num+=s[i]-'0';
            }
            else{
                res+=num*sign;
                num=0;
                if(s[i]=='+'){
                    sign=1;
                }
                if(s[i]=='-'){
                    sign=-1;
                }
                if(s[i]=='('){
                    nums.push(res);
                    nums.push(sign);
                    res=0;
                    sign=1;
                }
                if(s[i]==')'){
                    res*=nums.top();
                    nums.pop();
                    res+=nums.top();
                    nums.pop();
                }
            }
        }
        return res+=sign*num;
    }
};