679. 24 点游戏

https://leetcode-cn.com/problems/24-game/

你有 4 张写有 1 到 9 数字的牌。你需要判断是否能通过 *，/，+，-，(，) 的运算得到 24。
示例 1:

输入: [4, 1, 8, 7]
输出: True
解释: (8-4) * (7-1) = 24

---

思路：遍历,技巧：使用函数指针

enum class Oper{
    ADD,
    SUB,
    MUL,
    DEV
};

#define EQUALV 0.00001
#define RET 24

double calc(double v1,Oper op,double v2){
    switch(op){
        case Oper::ADD:
            return v1+v2;
        case Oper::SUB:
            return v1-v2;
        case Oper::MUL:
            return v1*v2;
        case Oper::DEV:
            if ( v2 != 0 ){
                return v1/v2;
            }
            return 0xffffff;
    }
    return 0xffffff;
}

double func1(int v1,Oper op1,int v2, Oper op2, int v3,Oper op3, int v4){//(a b) c d
    return calc(calc(calc(v1,op1,v2),op2,v3),op3,v4);
}

double func2(int v1,Oper op1,int v2, Oper op2, int v3,Oper op3, int v4){//(a b) (c d)
    return calc(calc(v1,op1,v2),op2,calc(v3,op3,v4));
}

double func3(int v1,Oper op1,int v2, Oper op2, int v3,Oper op3, int v4){//a (b c) d
    return calc(calc(v1,op1,calc(v2,op2,v3)),op3,v4);
}

double func4(int v1,Oper op1,int v2, Oper op2, int v3,Oper op3, int v4){//a ((b c) d)
    return calc(v1,op1,calc(calc(v2,op2,v3),op3,v4));
}

double func5(int v1,Oper op1,int v2, Oper op2, int v3,Oper op3, int v4){//a (b (c d))
    return calc(v1,op1,calc(v2,op2,calc(v3,op3,v4)));
}

class Solution {
public:
    bool judgePoint24(vector& nums) {
        static  int index[]={0,1,2,3,0,1,3,2,0,2,1,3,0,2,3,1,0,3,1,2,0,3,2,1,
            1,0,2,3,1,0,3,2,1,2,0,3,1,2,3,0,1,3,0,2,1,3,2,0,
            2,0,1,3,2,0,3,1,2,1,0,3,2,1,3,0,2,3,0,1,2,3,1,0,
            3,0,1,2,3,0,2,1,3,1,0,2,3,1,2,0,3,2,0,1,3,2,1,0};

        double (*fp[5])(int a,Oper op1,int b,Oper op2,int c,Oper op3, int d);
        fp[0] = func1;
        fp[1] = func2;
        fp[2] = func3;
        fp[3] = func4;
        fp[4] = func5;

        for ( Oper i1=Oper::ADD; i1<=Oper::DEV; i1=(Oper)((int)i1+1  ))
            for ( Oper i2=Oper::ADD; i2<=Oper::DEV; i2=(Oper)((int)i2+1  ))
                for ( Oper i3=Oper::ADD; i3<=Oper::DEV; i3=(Oper)((int)i3+1  ))
                    for ( int i=0; i<5; ++i  )
                        for ( int in=0; in<24; ++in  ){
                            double ret = (fp[i])(nums[index[in*4]],i1,nums[index[in*4+1]],i2,nums[index[in*4+2]],i3,nums[index[in*4+3]]) ;
                            if ( fabs(RET-ret) < EQUALV ){
                                return true;
                            }
                        }
        return false;
    }
};