Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers [start, end]. For each query, calculate the minimum number between index start and end in the given array, return the result list.

For array [1,2,7,8,5], and queries [(1,2),(0,4),(2,4)], return [2,1,5]

O(logN) time for each query

/**
 * Definition of Interval:
 * classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this->start = start;
 *         this->end = end;
 *     }
 */
class Solution { 
public:
    /**
     *@param A, queries: Given an integer array and an query list
     *@return: The result list
     */
    vector<int> intervalMinNumber(vector<int> &A, vector<Interval> &queries) {
        // write your code here
        vector<int> res;
        vector<int> dat = init(A);
        for (int i = 0; i < queries.size(); i++) {
            int min_val = query(dat, queries[i].start, queries[i].end + 1, 0, 0, dat.size() / 2);
            //注意區間是左閉右開；
            res.push_back(min_val);
        }
        return res;
        
    }
    vector<int> init (vector<int> &A) {
        int len = A.size();
        int n = 1;
        while ( n <= len) {
            n*=2;
        }
        vector<int> dat(2 * n, INT_MAX);
        for (int i = 0; i < A.size(); i++) {
            update(dat, i, A[i]);
        }
        // for (int i = 0; i < dat.size(); i++) {
        //     cout<<dat[i]<<" ";
        // }
        return dat;
    }
    void update(vector<int> &dat, int k, int value) {
        int n = dat.size() / 2;
        k += n - 1; //注意要去掉第一個；
        dat[k] = value;
        while (k > 0) {
            k = (k - 1) / 2;
            dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }
    int query (vector<int> &dat, int a, int b, int k, int l, int r) {
        // l 和 r 代表節點k所代表的區間
        //注意區間是左閉右開；
        if (r <= a || b <= l) {
            return INT_MAX;
        }
        if (a <= l && r <= b) {
            return dat[k];
        }else {
            int v1 = query(dat, a, b, k * 2 + 1, l, (l + r) / 2);
            int v2 = query(dat, a, b, k * 2 + 2, (l + r) / 2, r);
            return min(v1, v2);
        }
    }
};