227. Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

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​​

思路：

用一个栈来保存数字，

如果之前的运算符是+，则将该数压入栈；

如果之前的运算符是-，则将该数的负数压入栈；

如果之前的运算符是*，则弹出栈顶数，和当前的数相乘压入栈；

如果之前的运算符是-，则弹出栈顶数，和当前的数相除压入栈；

注意处理最后一次的情况

class Solution {
public:
  int calculate(string s) {
    int len = s.size();
    stack<int> stk;
    int num = 0;
    char sign = '+';
    for (int i = 0; i < len; i++){
      if (isdigit(s[i])){
        num = num * 10 + s[i] - '0';
      }
      if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/'||i==len-1){//不加else
        if (sign == '+'){
          stk.push(num);
        }
        else if (sign == '-'){
          stk.push(-num);
        }
        else if (sign == '*'){
          int top = stk.top();
          stk.pop();
          stk.push(top*num);
        }
        else{
          int top = stk.top();
          stk.pop();
          stk.push(top/num);
        }
        sign = s[i];
        num = 0;
      }
    }
    int res = 0;
    while (!stk.empty()){
      res += stk.top();
      stk.pop();
    }
    return res;
  }
};

224. Basic Calculator

Implement a basic calculator to evaluate a simple expression string.

​​(​​ 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​​

遇到 '(' 就把之前的结果和符号push进stack. 遇到')'就把 当前结果*stack中的符号 再加上stack中之前的结果.

class Solution {
public:
  int calculate(string s) {
    stack<int> stk;
    int len = s.size();

    int res = 0;
    int sign = 1;
    int i = 0;
    while (i<len){
      if (isdigit(s[i])){
        int num = 0;
        while (i < len&&isdigit(s[i])){
          num = num * 10 + s[i] - '0';
          i++;
        }
        res += num*sign;
      }
      else if (s[i] == '+'){
        sign = 1;
        i++;
      }
      else if (s[i] == '-'){
        sign = -1;
        i++;
      }
      else if (s[i] == '('){
        stk.push(res);
        stk.push(sign);
        res = 0;
        sign = 1;
        i++;
      }
      else if (s[i] == ')'){
        res = res*stk.top();
        stk.pop();
        res += stk.top();
        stk.pop();
        i++;
      }
      else{
        i++;
      }
    }
    return res;
  }
};