値を知っていれば、数値を「バケット」に入れて並べ替えることができます。それぞれの値に対して、バケットを作成し、そのバケットに反復処理を行うときにそのバケットに番号を追加します。
public class SortInBucket {
public static void main(String[] args) {
int[] x = {0,5,1,1,1,1,7,9,3,2,1,2,5,6};
System.out.println("Result of sorting: " + Arrays.toString(sortInBuckets(x)));
}
public static int[] sortInBuckets(int[] arr) {
List<List<Integer>> sortedNumbers = new ArrayList<>();
int[] sortedArr = new int[arr.length];
// create buckets 0 - 9
for (int i = 0; i < 10; i++) {
sortedNumbers.add(new ArrayList<>());
}
for (int i = 0; i < arr.length; i++) {
System.out.println("Found number " + arr[i] + " puting index " + i + " to bucket " + arr[i]);
sortedNumbers.get(arr[i]).add(i);
System.out.println("Bucket " + arr[i] + " is having " +sortedNumbers.get(arr[i]).size() + " numbers now.");
}
System.out.println();
System.out.println("The sortedNumbers (list with buckets) looks like following: " +sortedNumbers);
//just going through buckets and adding its numbers to sortedArr
int sortedIndex = 0;
for (List<Integer> bucket : sortedNumbers){
for (Integer num : bucket){
sortedArr[sortedIndex] = arr[num];
sortedIndex++;
}
}
return sortedArr;
}
}
以上持っているのコードを以下のように、すべての番号を持つあなたこれとは一度だけ、したがって、それは0-9から数字だけを持つ例えばO(n)
で行われ、あなたはそれを並べ替えることができますJFセバスチャンとSteve314で言及されたこの出力
Found number 0 puting index 0 to bucket 0
Bucket 0 is having 1 numbers now.
Found number 5 puting index 1 to bucket 5
Bucket 5 is having 1 numbers now.
Found number 1 puting index 2 to bucket 1
Bucket 1 is having 1 numbers now.
Found number 1 puting index 3 to bucket 1
Bucket 1 is having 2 numbers now.
Found number 1 puting index 4 to bucket 1
Bucket 1 is having 3 numbers now.
Found number 1 puting index 5 to bucket 1
Bucket 1 is having 4 numbers now.
Found number 7 puting index 6 to bucket 7
Bucket 7 is having 1 numbers now.
Found number 9 puting index 7 to bucket 9
Bucket 9 is having 1 numbers now.
Found number 3 puting index 8 to bucket 3
Bucket 3 is having 1 numbers now.
Found number 2 puting index 9 to bucket 2
Bucket 2 is having 1 numbers now.
Found number 1 puting index 10 to bucket 1
Bucket 1 is having 5 numbers now.
Found number 2 puting index 11 to bucket 2
Bucket 2 is having 2 numbers now.
Found number 5 puting index 12 to bucket 5
Bucket 5 is having 2 numbers now.
Found number 6 puting index 13 to bucket 6
Bucket 6 is having 1 numbers now.
The sortedNumbers (list with buckets) looks like following: [[0], [2, 3, 4, 5, 10], [9, 11], [8], [], [1, 12], [13], [6], [], [7]]
Result of sorting: [0, 1, 1, 1, 1, 1, 2, 2, 3, 5, 5, 6, 7, 9]
、トンを行うalghoritms彼はRadix sort(より一般化されたalghorithm)またはCounting sort( "強い"ではなく、よりシンプルでこの例では使用可能)と呼ばれています。
*最悪の場合、実行時に配列をソートすることができますO(n)*最悪の場合、リスト内の各項目を ' log(n) 'のうち –
しかし、仮定は一定時間で比較が行われると言うので、O(n)ではなくO(log n)であると言います – itaminul
これはどういう意味ですか? https://en.m.wikipedia.org/wiki/Counting_sort –