html5-img
1 / 7

O(n) Algorithms

O(n) Algorithms. Binsort/Bucketsort: Algorithm. /* Put a range of values in bins and then sort the contents of each bin and then output the values of each bin in order. */ void binsort( Elem[] a, int n ) { List[] B = new List[ MAX_KEY_VALUE ] for( i=0; i<n; i++ )

crescent
Télécharger la présentation

O(n) Algorithms

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. O(n) Algorithms

  2. Binsort/Bucketsort: Algorithm /* Put a range of values in bins and then sort the contents of each bin and then output the values of each bin in order. */ void binsort( Elem[] a, int n ) { List[] B = new List[ MAX_KEY_VALUE ] for( i=0; i<n; i++ ) B[ a[i].key ].append( a[i] ) int index = 0 for( i=0; i< MAX_KEY_VALUE; i++) { for( B[i].first(); B[i].isInList(); B[i].next() ) { a[i] = B[i].currentValue() } } }

  3. 0 1 2 3 4 5 6 7 8 9 Bins: 0 1 2 3 4 5 6 7 8 9 Simple Binsort 8 5 6 0 3 9 2 4 1 7 A[]: 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 for( int i = 0; i < n; i++ ) { a[i].key = bin[i] } for( int i = 0; i < n; i++ ) { bin[ a[i].key ] = a[i].key }

  4. 0 3 5 6 9 Bins: 0 1 2 3 4 5 6 7 8 9 Simple Binsort (another example) 9 5 6 0 3 A[]: 0 1 2 3 4 0 3 5 6 9 for( int i = 0, j=0; i < n; i++ ) { if( bin[i].isNotEmpty() ) { a[j].key = bin[i] j++ } } for( int i = 0; i < n; i++ ) { bin[ a[i].key ] = a[i].key }

  5. 5 8 5 6 8 45 6 39 8 42 39 39 42 5 42 45 45 6 75 75 75 91 97 91 95 97 95 91 97 95 8 8 5 6 8 8 5 6 8 5 6 8 6 6 6 39 0 0 39 39 0 3 42 45 42 45 45 42 3 3 9 9 9 2 2 2 4 4 75 75 75 4 1 1 1 7 91 95 97 97 91 95 7 91 95 97 7 buckets: 0 1 2 3 4 5 6 7 8 9 BucketSort (buckets by 10’s) Keys: 0 4 3 4 0 0 7 9 9 9 A[]: for each Bucket { sort the elements in it. } for each element in A[] { put the element in its proper bucket B[a[].key] } for each Bucket { copy the elements back into the original array }

  6. Radixsort: Algorithm void radixsort( Elem[] a, Elem[] B, int n, int k, int r, Elem[] count) { // count[i] stores the number of records in bin[i] for( int i=0, rtok=1; i<k; i++, rtok*=r) //for k digits { for( int j=0; j<r; j++) count[j]=0; //initialize count //Count the no. of records for each bin on this pass for( j=0; j<n; j++ ) count[ (a[j].key / rtok) % r ] ++ // Index B: count [j] will be index for last slot of bin j for( j=1; j<r; j++ ) count[j] = count[j-1] + count[j] // put records into bins working from bottom of each bin // since bins fill from the bottom, j counts downwards for( j=n-1; j>=0; j--) B[--count[ (a[j].key / rtok) % r ] ] = a[j] for( j=0; j<n; j++ ) a[j] = B[j] //copy B back into a } }

  7. 5 91 8 5 6 45 6 42 8 45 39 8 5 39 42 39 75 42 42 5 45 45 95 6 75 75 75 6 91 97 97 91 8 95 95 91 97 95 97 39 Sorted: 2nd Pass: 1st Pass: Original: Radixsort: Illustration Sort the numbers according to the ones place. Sort the numbers according to the tens place.

More Related