Bucket Sort

(referrence: GeekforGeeks)

Bucket sort is mainly useful when input is uniformly distributed over a range. For example, consider the following problem. 
Sort a large set of floating point numbers which are in range from 0.0 to 1.0 and are uniformly distributed across the range. How do we sort the numbers efficiently?

The lower bound for Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick Sort, etc) is O(n log n).

And, counting sort cannot be applied here as we use keys as index in counting sort.

bucketSort(arr[], n)
1) Create n empty buckets (Or lists).
2) Do following for every array element arr[i].
.......a) Insert arr[i] into bucket[n*array[i]]
3) Sort individual buckets using insertion sort.
4) Concatenate all sorted buckets.

If we assume that insertion in a bucket takes O(1) time then steps 1 and 2 of the above algorithm clearly take O(n) time.

The O(1) is easily possible if we use a linked list to represent a bucket.

Step 4 also takes O(n) time as there will be n items in all buckets.

The main step to analyze is step 3. This step also takes O(n) time on average if all numbers are uniformly distributed.

 1 class Solution {
 2     public static void bucketSort(int[] arr, int n) {
 3         // Create n empty buckets
 4         List<Integer>[] buckets = new ArrayList[n];
 5         for (int i = 0; i < n; i++)
 6             buckets[i] = new ArrayList<Integer>();
 7 
 8         for (int i = 0; i < arr.length; i++) {
 9             int index = n * arr[i];
10             buckets[index].add(arr[i]);
11         }
12 
13         // Sort buckets
14         for (int i = 0; i < n; i++)
15             Collections.sort(buckets[i]);
16         // Connect buckets
17         int p = 0;
18         for (int i = 0; i < n; i++) {
19             for (int j = 0; j < buckets[i].size(); j++) {
20                 arr[index++] = b[i].get(j);
21             }
22         }
23     }
24 }

 

转载于:https://www.cnblogs.com/ireneyanglan/p/4858174.html