Count of Smaller Number

最新推荐文章于 2024-08-06 18:00:00 发布

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最新推荐文章于 2024-08-06 18:00:00 发布

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原文链接：http://www.cnblogs.com/joycelee/p/4516165.html

本文介绍了一种算法解决方案，通过快速排序和二分查找来计算数组中小于特定查询值的元素数量。该方法适用于整数数组，并提供了一个具体的Java实现案例。

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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

 1 public class Smaller {
 2     
 3         public ArrayList<Integer> countOfSmallerNumber(int[] A, int[] queries) {
 4             ArrayList<Integer> list = new ArrayList<Integer>();
 5             if (queries == null || queries.length == 0)
 6                 return list;
 7             if(A == null || A.length == 0) {
 8                 for (int i = 0; i < queries.length; i++) {
 9                     list.add(0);
10                 }
11                 return list;
12             }
13 
14             sort(A, 0, A.length-1);
15             
16             for (int i = 0; i < queries.length; i++) {
17                 list.add(binary(A, queries[i]));
18             }
19             System.out.println(list);
20             return list;
21         }
22         
23         public int binary(int[] A, int k) {
24             int i = 0, j = A.length-1;
25             while (i <= j) {
26                 int mid = (j - i) / 2 + i;
27                 if (A[mid] >= k) j = mid - 1;
28                 else i = mid + 1;
29             }
30             return i;
31         }
32         
33         public void sort(int[] A, int start, int end) {
34             if (start >= end) return;
35             int i = start, j = end;
36             int tmp = A[start];
37             while (i < j) {
38                 while (i < j && A[j] >= tmp) j--;
39                 A[i] = A[j];
40                 while (i < j && A[i] < tmp) i++;
41                 A[j] = A[i];
42             }
43             A[i] = tmp;
44             sort(A, start, i-1);
45             sort(A, i+1, end);
46         }
47         public static void main(String[] args) {
48             Smaller sol = new Smaller();
49             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};
50             int[] queries = {10,43,2,17,28,75,75,12};
51             sol.countOfSmallerNumber(A, queries);
52         }
53     }