leetcode 16.26 计算器 C++解法

这个代码是参考了评论区的题解，一变读取一遍计算，使用prev_op来实现减号的变化。

比之前的中缀转后缀再计算要快得多

这个是之前的版本https://blog.youkuaiyun.com/dyyzlzc/article/details/104337379

class Solution
{
public:
    int calculate(string s)
    {
        s+="#";
        char op = '+';
        list<int> nums;
        int *num = nullptr;
        for (auto ch : s)
        {
            if (ch == ' ')
                continue;
            if (ch <= '9' && ch >= '0')
            {
                if (num == nullptr)
                {
                    num=new int;
                    (*num) = ch - '0';
                }
                else
                {
                    (*num) = (*num) * 10 + (ch - '0');
                }
                continue;
            }
            if (ch == '+' || ch == '-' || ch == '*' || ch == '/' || ch=='#')
            {
                if (op == '+')
                {
                    nums.push_back(*num);
                }
                else if (op == '-')
                {
                    nums.push_back(-(*num));
                }
                else if (op == '*')
                {
                    int r = nums.back();
                    nums.pop_back();
                    r *= *num;
                    nums.push_back(r);
                }
                else if(op=='/')
                {
                    int r = nums.back();
                    nums.pop_back();
                    r /= *num;
                    nums.push_back(r);
                }else{
                    nums.push_back(*num);
                    break;
                }
                delete num;
                num=nullptr;
                op=ch;
            }
        }
        return accumulate(nums.begin(),nums.end(),0);
    }

};