Count of Smaller Number

Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) and an query list. For each query, give you an integer, return the number of element in the array that are smaller that the given integer.

hint: 

Solution 1

　　Quick Sort + Binary Search

Solution 2

　　Segment Tree

public class Smaller {
    
        public ArrayList<Integer> countOfSmallerNumber(int[] A, int[] queries) {
            ArrayList<Integer> list = new ArrayList<Integer>();
            if (queries == null || queries.length == 0)
                return list;
            if(A == null || A.length == 0) {
                for (int i = 0; i < queries.length; i++) {
                    list.add(0);
                }
                return list;
            }

            sort(A, 0, A.length-1);
            
            for (int i = 0; i < queries.length; i++) {
                list.add(binary(A, queries[i]));
            }
            System.out.println(list);
            return list;
        }
        
        public int binary(int[] A, int k) {
            int i = 0, j = A.length-1;
            while (i <= j) {
                int mid = (j - i) / 2 + i;
                if (A[mid] >= k) j = mid - 1;
                else i = mid + 1;
            }
            return i;
        }
        
        public void sort(int[] A, int start, int end) {
            if (start >= end) return;
            int i = start, j = end;
            int tmp = A[start];
            while (i < j) {
                while (i < j && A[j] >= tmp) j--;
                A[i] = A[j];
                while (i < j && A[i] < tmp) i++;
                A[j] = A[i];
            }
            A[i] = tmp;
            sort(A, start, i-1);
            sort(A, i+1, end);
        }
        public static void main(String[] args) {
            Smaller sol = new Smaller();
            int[] A = {55,81,56,91,35,92,10,53,27,94,64,45,19,44,52,19,79,12,16,90,97,33,73,2,20,68,19,7,17,62,45,48,62,26,85,4,63,67,56,16};
            int[] queries = {10,43,2,17,28,75,75,12};
            sol.countOfSmallerNumber(A, queries);
        }
    }