Re: [閒聊] 每日leetcode

[oin1104 (是oin的說)]

題目:
給你字串
-1/2+1/2
3/6-1/3+2/5-10/2
要你算出結果

思路:
機掰
超麻煩
直接做一個class num
然後用operator處理加減
生成num就暴力的一個一個字放進去

姆咪

```cpp

class num{
public:

    int up;
    int down;

    num(){
        up = 0;
        down = 1;
    }

    num operator+(const num &a){
        num res;
        res.up = up * a.down + a.up * down;
        res.down = down * a.down;
        for(int i = 2 ; i <= 100 ; i ++)
        {
            while( res.up%i == 0 && res.down%i == 0 )
            {
                res.up /= i;
                res.down /= i;
            }
        }
        return res;
    }
};


class Solution {
public:
    string fractionAddition(string exp)
    {

        int len = exp.size();
        int i = 0 ;
        num k ;
        int kneg = 1;
        if (exp[i] == '-'){
            kneg = -1;
            i++;
        }

        k.up = exp[i] - '0';
        if (i + 1 < len && exp[i + 1] == '0') {
            k.up = 10;
            i++;
        }
        k.up *= kneg;

        if (i + 3 < len && exp[i + 2] == '1' && exp[i + 3] == '0')
        {
            k.down = 10;
            i += 4;
        }
        else
        {
            k.down = exp[i + 2] - '0';
            i += 3;
        }

        while (i < len) {
            num now;
            int neg = 1;

            if (exp[i] == '-')
            {
                neg = -1;
                i++;
            }
            else if (exp[i] == '+')
            {
                i++;
            }

            now.up = exp[i] - '0';
            if (i + 1 < len && exp[i + 1] == '0')
            {
                now.up = 10;
                i++;
            }
            now.up *= neg;

            if (i + 3 < len && exp[i + 2] == '1' && exp[i + 3] == '0')
            {
                now.down = 10;
                i += 4;
            }
            else
            {
                now.down = exp[i + 2] - '0';
                i += 3;
            }

            k = k + now;
        }

        string ress;
        ress += to_string(k.up);
        ress += "/";
        ress += to_string(k.down);

        return ress;
    }
};```



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