leetcode 679. 24 Game

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本文探讨了如何通过加减乘除操作使四个1至9之间的数字等于24的问题。采用深度优先搜索策略，利用分数结构处理除法运算，确保所有可能的操作组合都被考虑。示例包括成功的解决方案和不成功的案例。

679. 24 Game

You have 4 cards each containing a number from 1 to 9. You need to judge whether they could operated through *, /, +, -, (, )to get the value of 24.

Example 1:

Input: [4, 1, 8, 7]
Output: True
Explanation: (8-4) * (7-1) = 24

Example 2:

Input: [1, 2, 1, 2]
Output: False

Note:

The division operator / represents real division, not integer division. For example, 4 / (1 - 2/3) = 12.
Every operation done is between two numbers. In particular, we cannot use - as a unary operator. For example, with [1, 1, 1, 1] as input, the expression -1 - 1 - 1 - 1 is not allowed.
You cannot concatenate numbers together. For example, if the input is [1, 2, 1, 2], we cannot write this as 12 + 12.

题目：每个数字都必须用到，且只能用一次。

1、DFS。

2、取出数组中两个数，然后相互加减乘除。

3、为了处理除法，所以用一个shu结构来存储分母分子。

class shu{
public:
    int fenzi;
    int fenmu;
    shu(int x, int y = 1):fenzi(x), fenmu(y){}
};

class Solution {
public:
    bool judgePoint24(vector<int>& nums)
    {
        vector<shu> num;
        for (auto it : nums)
            num.push_back(shu(it));
        return helper(num);
    }
    
    bool helper(vector<shu>& nums)  //成功返回1
    {
        if (nums.size() == 1 && nums[0].fenmu != 0 && nums[0].fenzi % nums[0].fenmu == 0 && nums[0].fenzi / nums[0].fenmu == 24)
            return true;
        
        int size = nums.size();
        for (int i = 0; i < size; i++)
        {
            for (int j = i + 1; j < size; j++)
            {
                vector<shu> tmp_p = nums;
                tmp_p.erase(tmp_p.begin() + i);
                tmp_p.erase(tmp_p.begin() + j - 1);
                tmp_p.push_back( shu(nums[i].fenzi*nums[j].fenmu + nums[i].fenmu * nums[j].fenzi, nums[i].fenmu * nums[j].fenmu));
                if (helper(tmp_p)) return true;
                
                vector<shu> tmp_j = nums;
                tmp_j.erase(tmp_j.begin() + i);
                tmp_j.erase(tmp_j.begin() + j - 1);
                tmp_j.push_back( shu(abs(nums[i].fenzi * nums[j].fenmu - nums[i].fenmu * nums[j].fenzi), nums[i].fenmu * nums[j].fenmu));
                if (helper(tmp_j)) return true;
                
                vector<shu> tmp_c = nums;
                tmp_c.erase(tmp_c.begin() + i);
                tmp_c.erase(tmp_c.begin() + j - 1);
                tmp_c.push_back( shu(nums[i].fenzi * nums[j].fenzi, nums[i].fenmu * nums[j].fenmu));
                if (helper(tmp_c)) return true;

                vector<shu> tmp_d1 = nums;
                tmp_d1.erase(tmp_d1.begin() + i);
                tmp_d1.erase(tmp_d1.begin() + j - 1);
                tmp_d1.push_back( shu(nums[i].fenzi * nums[j].fenmu, nums[i].fenmu * nums[j].fenzi));
                if (helper(tmp_d1)) return true;
                
                vector<shu> tmp_d2 = nums;
                tmp_d2.erase(tmp_d2.begin() + i);
                tmp_d2.erase(tmp_d2.begin() + j - 1);
                tmp_d2.push_back( shu(nums[i].fenmu * nums[j].fenzi, nums[i].fenzi * nums[j].fenmu));
                if (helper(tmp_d2))  return true;
            }
        }
        return false;
    }
};