Data Structure (Heap Sort)
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This file explains about heap sort in programming. This file was used in my Data Structure class.
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Data Structure (Heap Sort)
- 3. Heap Heap Tree Complete Binary Tree (CBT) Max Heap Value of node >= value of its child Min Heap Value of node <= value of its child
- 4. Example of Heap Max Heap Min Heap 24 19 13 15 11 5 8 4 4 5 13 8 11 19 15 24 k 2k 2k+1 2k+1 2k+12k 2k 2k
- 6. • Making of heap • Heap sort Processes in Heap
- 7. Making of Heap • Shift down from middle node (Sum of node/2) until first node 1 2 3 4 14 11 57 5 6 3 2 1 2 3 54 6 • N = 6, Middle = N/2 = 6/2 = 3 • Reorganize on third node • Reorganize on second node • Reorganize on first node 1 2 3 4 14 11 5 7 5 6 3 2 7 3 14 11 2 53 14 2 57 14 11 7
- 8. Heap Sort 1 a. Binary tree in Max heap state. b. “Fired” root and swap with last node. c. Subtract number of nodes with 1. d. If N > 1 then reorganize heap again. e. Repeat step b to d until node is empty (N=0) 1 2 3 54 6 1 2 3 4 14 11 5 7 5 6 3 2142 14 3 2 5 14 11 7 2
- 9. Heap Sort 1 a.Binary tree in Max heap state. b.“Fired” root and swap with last node. c. Subtract number of nodes with 1. d.If N > 1 then reorganize heap again. e.Repeat step b to d until node is empty (N=0) 1 2 3 54 6 1 2 3 4 14 11 5 7 5 6 3 2142 14 3 2 5 14 11 7 2
- 10. Heap Sort 2 Reorganize heap Middle = N/2 = 5/2 = 2 b. “Fired” root and swap with last node. c. Subtract number of nodes with 1. d. If N > 1 then reorganize heap again. e. Repeat step b to d until node is empty (N=0) Reorganize heap on second node. Reorganize heap on first node. 1 2 3 54 3 511 7 2 1 2 3 4 11 5 7 5 6 3 2142 14 11 27 2 2 11 27 2112 11
- 11. Heap Sort 3 Reorganize heap Middle = N/2 = 4/2 = 2 b. “Fired” root and swap with last node. c. Subtract number of nodes with 1. d. If N > 1 then reorganize heap again. e. Repeat step b to d until node is empty (N=0) Reorganize heap on second node. Reorganize heap on first node. 1 2 3 4 3 5 2 7 1 2 3 4 5 7 5 6 3 214147 2112 11 2 7 7 2 3 2 3 2 2 72 7
- 12. Heap Sort 4 Reorganize heap Middle = N/2 = 3/2 = 1 b. “Fired” root and swap with last node. c. Subtract number of nodes with 1. d. If N > 1 then reorganize heap again. e. Repeat step b to d until node is empty (N=0) Reorganize heap on first node. 1 2 3 5 2 3 1 2 3 4 5 7 5 6 3 21414211113 272 7 5 2 5 2 2 52 5
- 13. Heap Sort 5 Reorganize heap Middle = N/2 = 2/2 = 1 b. “Fired” root and swap with last node. c. Subtract number of nodes with 1. d. If N > 1 then reorganize heap again. e. Repeat step b to d until node is empty (N=0) Reorganize heap on first node. 1 2 2 3 1 2 3 4 5 7 5 6 3 21414211113 277252 5 2 3 3 2 2 32 3
- 14. Heap Sort 6 Because N = 1 then reorganize isn’t happen. b. “Fired” root c. Subtract number of nodes with 1. Because N = 0 then sorting processes is finish. 1 2 1 2 3 4 5 7 5 6 3 21414211113 277252 52332
- 15. m CASE
- 16. Node have 2 Case Example Sort these name using heap sort method in descending way. Num. Name 1 Rahmat 2 Didin 3 Ahmad 4 Joned 5 Syahrul 6 Riki 7 Arif 8 Susi 9 Donni 10 Asih
- 17. Node have 2 Making of CBT Rahmat AhmadDidin Joned Syahrul Riki Arif Susi Donni Asih Not Heap Complete Binary Tree Rahmat Didin Ahmad Joned Syahrul Riki Arif Susi Donni Asih 1 2 3 4 5 6 7 8 9 10
- 18. Node have 2 Heap Sort 1 1 2 3 4 5 6 7 8 9 10 Rahmat AhmadDidin Joned Syahrul Riki Arif Susi Donni Asih Heap Syahrul Asih Joned Donni Didin Asih Rahmat Ahmad Arif Rahmat Ahmad Asih Arif Donni Didin Riki Rahmat Susi Joned Syahrul
- 19. Node have 2 Heap Sort 2 Riki Susi Not Heap SyahrulJoned Donni Didin Asih Ahmad Arif Rahmat Syahrul 1 2 3 4 5 6 7 8 9 10 Ahmad Asih Arif Donni Didin Riki Rahmat Susi Joned SyahrulAhmadSyahrul Ahmad
- 20. Node have 2 Heap Sort 3 Riki Susi Heap Joned Donni Didin Asih Arif Rahmat SyahrulArif SyahrulRahmat Syahrul 1 2 3 4 5 6 7 8 9 10 Asih Arif Donni Didin Riki Rahmat Susi Joned SyahrulAhmadSyahrul AhmadArif SyahrulRahmat Syahrul
- 21. Node have 2 Heap Sort 4 Riki Susi Not Heap Joned Donni Didin Asih Arif Rahmat Syahrul Joned 1 2 3 4 5 6 7 8 9 10 Asih Donni Didin Riki Susi Joned SyahrulAhmadAhmadArif Rahmat SyahrulJoned ArifArif
- 22. Node have 2 Heap Sort 5 Riki Susi Heap Donni Didin Asih Rahmat Syahrul JonedAsih JonedDidin Joned 1 2 3 4 5 6 7 8 9 10 Asih Donni Didin Riki Susi Joned SyahrulAhmadAhmadRahmat SyahrulJoned ArifArifJonedAsih Didin Joned
- 23. Node have 2 Heap Sort 6 Riki Susi Not Heap Donni Rahmat Syahrul Asih Didin Joned Susi 1 2 3 4 5 6 7 8 9 10 Donni Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifAsih Didin JonedSusi AsihAsih
- 24. Node have 2 Heap Sort 7 Riki Heap Donni Rahmat Syahrul Didin Joned SusiDidin SusiDonni Susi 1 2 3 4 5 6 7 8 9 10 Donni Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifDidin JonedSusi AsihAsihDidin SusiDonni Susi
- 25. Node have 2 Heap Sort 8 Riki Not Heap Rahmat SyahrulJoned Didin Donni Susi Syahrul 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifJoned AsihAsihDidin Donni SusiSyahrul DidinDidin
- 26. Node have 2 Heap Sort 9 Riki Heap Rahmat Joned Donni Susi SyahrulDonni SyahrulJoned Syahrul 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifJoned AsihAsihDonni SusiSyahrul DidinDidinDonni SyahrulJoned Syahrul
- 27. Node have 2 Heap Sort 10 Riki Not Heap Rahmat Susi Donni Joned Syahrul Riki 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifAsihAsihSusi DidinDidinDonni Joned SyahrulRiki DonniDonni
- 28. Node have 2 Heap Sort 11 Heap Rahmat Susi Joned Syahrul RikiJoned Riki 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifAsihAsihSusi DidinDidinJoned SyahrulRiki DonniDonniJoned Riki
- 29. Node have 2 Heap Sort 12 Not Heap Rahmat Susi Syahrul Joned Riki Syahrul 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJoned RikiSyahrul JonedJoned
- 30. Node have 2 Heap Sort 13 Heap Rahmat Susi Riki SyahrulRahmat Syahrul 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadRahmat Syahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniSyahrul JonedRiki DonniRahmat Syahrul
- 31. Node have 2 Heap Sort 14 Not Heap Susi Riki Rahmat Syahrul Susi 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadSyahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJonedRiki DonniRahmat SyahrulSusi RahmatRahmat
- 32. Node have 2 Heap Sort 15 Heap Riki Syahrul SusiRiki Susi 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadSyahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJonedRiki DonniSyahrulSusi RahmatRahmatRiki Susi
- 33. Node have 2 Heap Sort 17 Heap Susi Syahrul 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadSyahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJonedDonniSyahrul RahmatRahmatSyahrul Susi Riki Susi Syahrul Susi Syahrul
- 34. Node have 2 Heap Sort 16 Not Heap Syahrul Riki Susi 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadSyahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJonedDonniSyahrul RahmatRahmatRiki Susi Riki Syahrul Syahrul Riki
- 35. Node have 2 Heap Sort 18 Heap 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadSyahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJonedDonniSyahrul RahmatRahmatRiki Susi Syahrul Susi Syahrul Syahrul SusiSyahrul Susi
- 36. Node have 2 Heap Sort 19 Syahrul 1 2 3 4 5 6 7 8 9 10 Riki Susi Joned SyahrulAhmadAhmadSyahrul ArifArifAsihAsihSusi DidinDidinSyahrul DonniDonniJonedDonniSyahrul RahmatRahmatRikiSyahrulSusiSyahrul SusiSyahrul
- 37. Contact Person: Adam Mukharil Bachtiar Informatics Engineering UNIKOM Jalan Dipati Ukur Nomor. 112-114 Bandung 40132 Email: adfbipotter@gmail.com Blog: http://adfbipotter.wordpress.com Copyright © Adam Mukharil Bachtiar 2012