leetcode1906 查询差绝对值的最小值

思路：

注意到数组a中元素的取值范围只有1-100，可以通过枚举1-100中每个数字是否出现的方式来计算。具体可以通过前缀和实现。

实现：

class Solution
{
public:
    vector<int> minDifference(vector<int>& nums, vector<vector<int>>& queries)
    {
        vector<int>res;
        int n = nums.size();
        vector<vector<int>> sum(n + 1, vector<int>(101, 0));
        for (int i = 0; i < n; i++)
        {
            for (int j = 1; j <= 100; j++)
            {
                sum[i + 1][j] = sum[i][j];
            }
            sum[i + 1][nums[i]]++;
        }
        for (int i = 0; i < queries.size(); i++)
        {
            int l = queries[i][0], r = queries[i][1];
            vector<int> v;
            for (int j = 1; j <= 100; j++)
            {
                int x = sum[r + 1][j] - sum[l][j];
                if (x) v.push_back(j);
            }
            int minn = 100;
            for (int j = 0; j < v.size() - 1; j++)
            {
                minn = min(minn, v[j + 1] - v[j]);
            }
            if (minn == 100) res.push_back(-1);
            else res.push_back(minn);
        }
        return res;
    }
}