Basic Calculator,Basic Calculator II

一.Basic Calculator

Total Accepted: 18480 Total Submissions: 94750 Difficulty: Medium

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.

 (M) Evaluate Reverse Polish Notation (M) Basic Calculator II (M) Different Ways to Add Parentheses (H) Expression Add Operators

 

/*
已知条件：
1.只包含空格,+,-,(,),非负整数
2.假设输入一定合法
测试用例
"6-(4-9)+7"
"8+(1+(4+5-(7-3)-2)-3)+(6+8)"
"1 + 1+2"
用括号分割表达式，遇到()先算括号表达式内容
*/
class Solution {
public:
    int calculate(string &s ,int& start,int end)
    {
        char pre_op='+';
        int num=0,res=0;
        while(start<end){
            if(s[start]==' '){
                start++;continue;
            }
            if(isdigit(s[start])){
                num = num*10+(s[start++]-'0');
            }else if(s[start]=='('){
                num = calculate(s,++start,end);
                start++;
            }else if(s[start]==')'){
                return pre_op=='+' ? res+num:res-num;
            }else{
                res = pre_op=='+' ? res+num:res-num;
                pre_op = s[start++];
                num = 0;
            }
        }
        return pre_op=='+' ? res+num:res-num;
    }
    int calculate(string s) {
        int start = 0;
        return calculate(s,start,s.size());
    }
};

Next challenges: (M) Evaluate Reverse Polish Notation (M) Different Ways to Add Parentheses (H) Expression Add Operators

 

二.Basic CalculatorII

Total Accepted: 14291 Total Submissions: 64507 Difficulty: Medium

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.

 (M) Basic Calculator (H) Expression Add Operators

对表达式按+-划分，含乘除的表达式部分当做一个整体，如果当前运算符是+-号，说明前一个表达的结果已经计算完成，那么把前一个表达式的结果加到输出结果中，如果是*/则说明表达式尚未结束。

 

/*
题目已知：
1.只包含非负整数，加减乘除,空格
2.假设输入一直合法
涉及的几个点：
1.数字分割
2.字符串转整数
3.运算符的优先级
可能隐藏的点：
大数
测试案例：
"0"
"1+0"
"1+10"
" 13 + 24 "
" 23+ 34*3 /2 "
" 12 - 7*3/2 + 35/7-3 "
" 7*2/3 + 9"
"12 - 7*3/2 + 35/7"
*/
class Solution {
public:
    int calculate(string s) {
        int size = s.size();
        long long int exp_res = 0,res=0;
        long int num =0;
        char pre_op='+';
        for(int i=0;i<size;i++){
            if(s[i]==' ') continue;
            if(isdigit(s[i])){
                num = num*10+(s[i]-'0');
                if(i+1 == size || !isdigit(s[i+1])){//如果下一个位置是最后一个位置，或者下一个位置不是数字了
                    if(pre_op=='+'){
                        exp_res = num;
                    }else if(pre_op=='-'){
                        exp_res = -num;
                    }else if(pre_op=='*'){
                        exp_res *= num;
                    }else{
                        exp_res /= num;
                    }
                }
            }else{
                if(s[i]=='+' || s[i]=='-'){
                    res += exp_res;
                }
                pre_op = s[i];
                num=0;
            }
        }
        return res+exp_res;
    }
};

Next challenges: (M) Basic Calculator (H) Expression Add Operators