LeetCode 1755. 最接近目标值的子序列和（状态枚举 + 双指针）

示例 1：
输入：nums = [5,-7,3,5], goal = 6
输出：0
解释：选择整个数组作为选出的子序列，元素和为 6 。
子序列和与目标值相等，所以绝对差为 0 。

示例 2：
输入：nums = [7,-9,15,-2], goal = -5
输出：1
解释：选出子序列 [7,-9,-2] ，元素和为 -4 。
绝对差为 abs(-4 - (-5)) = abs(1) = 1 ，是可能的最小值。

示例 3：
输入：nums = [1,2,3], goal = -7
输出：7
 
提示：
1 <= nums.length <= 40
-10^7 <= nums[i] <= 10^7
-10^9 <= goal <= 10^9

class Solution {
public:
    int minAbsDifference(vector<int>& nums, int goal) {
        int n = nums.size();
        vector<int> arr1, arr2;
        getsum(nums, 0, n/2, arr1);
        getsum(nums, n/2, n, arr2);
        int i = 0, j = arr2.size()-1, n1 = arr1.size();
        int diff = INT_MAX, sum;
        while(i < n1 && j >= 0)
        {
            sum = arr1[i] + arr2[j];
            diff = min(diff, abs(sum-goal));
            if(sum > goal)
                j--;
            else if(sum < goal)
                i++;
            else
                break;
        }
        return diff;
    }
    void getsum(vector<int>& nums, int l, int r, vector<int>& arr)
    {
        int n = r-l;
        arr.resize(1<<n);
        for(int i = 0; i < (1<<n); i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(i & (1 << j))
                    continue;// i 状态 包含 j 数字
                arr[i+(1<<j)] = arr[i] + nums[l+j];
            }
        }
        sort(arr.begin(), arr.end());
    }
};