LeetCode: 227. Basic Calculator II

LeetCode: 227. Basic Calculator II

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.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

• You may assume that the given expression is always valid.
• Do not use the ​​eval​​ built-in library function.

解题思路

将单个数字和乘除法作为加法运算的 “项(item)”，每次获取下一项进行计算。

AC 代码

class Solution {
private:
    int Op(char op, int lhv, int rhv)
    {
        if(op == '+') return lhv+rhv;
        else if(op == '-') return lhv-rhv;
        else if(op == '*') return lhv*rhv;
        else return lhv/rhv;
    }
    // idx, result
    pair<int, int> nextItem(string str, int beg)
    {
        int result = 0;
        int curNum = 0;
        char op = '+';
        size_t i;

        size_t pos = str.find_first_of("+-", beg);
        if(pos == str.npos) pos = str.size();
        for(i = beg; i <= pos; ++i)
        {
            if(str[i] == ' ') continue;
            if(i == pos)
            {
                result = Op(op, result, curNum);
                break;
            }
            if(str[i] >= '0' && str[i] <= '9')
            {
                curNum = curNum*10+str[i]-'0';
            }
            else if(str[i] == '*' || str[i] == '/')
            {
                result = Op(op, result, curNum);
                op = str[i];
                curNum = 0;
            }
        }

        return {i, result};
    }
public:
    int calculate(string s) {
        int result = 0;
        int curNum = 0;
        int idx = 0;
        char op = '+';
        pair<int, int> curPair{0,0};

        while(idx < s.size())
        {
            if(s[idx] == ' ') 
            {
                ++idx;
                continue;
            }

            if(s[idx] =='+' || s[idx] =='-')
            {
                op = s[idx++];
                continue;
            }

            curPair = nextItem(s, idx);
            result = Op(op, result, curPair.second);
            idx = curPair.first;
        }

        return result;
    }
};