1 00:00:00,500 --> 00:00:02,840 [Powered by Google Translate] [Bubble aina] 2 00:00:02,840 --> 00:00:04,560 [JACKSON STEINKAMP HARVARD UNIVERSITY] 3 00:00:04,560 --> 00:00:07,500 [HII NI CS50. CS50TV] 4 00:00:08,000 --> 00:00:11,730 Bubble Panga ni mfano wa algorithm kuchagua - 5 00:00:11,730 --> 00:00:14,460 kwamba ni, utaratibu kwa ajili ya kuchagua seti ya vipengele katika 6 00:00:14,460 --> 00:00:15,840 wakipanda au kushuka ili. 7 00:00:15,840 --> 00:00:18,710 Kwa mfano, kama alitaka kutatua safu yenye namba 8 00:00:18,710 --> 00:00:23,060 [3, 5, 2, 9], utekelezaji sahihi wa aina Bubble atarudi 9 00:00:23,060 --> 00:00:26,260 Iliyopangwa safu [2, 3, 5, 9] katika wakipanda ili. 10 00:00:26,260 --> 00:00:28,850 Sasa, mimi naenda kueleza katika pseudocode jinsi algorithm kazi. 11 00:00:28,850 --> 00:00:34,000 >> Hebu sema tuko kuchagua orodha ya integers 5-3, 2, 9, 6, na 5. 12 00:00:34,000 --> 00:00:37,650 algorithm kuanza kwa kuangalia mambo mawili ya kwanza, 3 na 2, 13 00:00:37,650 --> 00:00:40,850 na kuangalia kama uko nje ya utaratibu jamaa na kila mmoja. 14 00:00:40,850 --> 00:00:43,150 Wao ni - 3, ni mkubwa kuliko 2. 15 00:00:43,150 --> 00:00:45,190 Kuwa ili wakipanda, wanapaswa kuwa na njia nyingine kote. 16 00:00:45,190 --> 00:00:46,610 Hivyo, sisi byta yao. 17 00:00:46,610 --> 00:00:49,760 Sasa orodha inaonekana kama hii: [2, 3, 9, 6, 5]. 18 00:00:49,760 --> 00:00:52,450 >> Next, sisi kuangalia vipengele pili na wa tatu, 3 na 9. 19 00:00:52,450 --> 00:00:55,770 Wao ni katika mpangilio sahihi jamaa na kila mmoja. 20 00:00:55,770 --> 00:00:58,800 Hiyo ni, 3 ni chini ya 9 hivyo algorithm haina byta yao. 21 00:00:58,800 --> 00:01:01,900 Next, sisi kuangalia 9 na 6. Wao ni nje ya utaratibu. 22 00:01:01,900 --> 00:01:04,260 >> Hivyo, tunahitaji wabadilishane yao kwa sababu 9, ni mkubwa kuliko 6. 23 00:01:04,260 --> 00:01:08,840 Mwisho, sisi kuangalia integers miwili iliyopita, 9 na 5. 24 00:01:08,840 --> 00:01:10,850 Wao ni nje ya utaratibu, hivyo ni lazima swapped. 25 00:01:10,850 --> 00:01:13,360 Baada ya kwanza kupita kamili kupitia orodha, 26 00:01:13,360 --> 00:01:17,140 inaonekana kama hii: [2, 3, 6, 5, 9]. 27 00:01:17,140 --> 00:01:19,690 Si mbaya. Ni karibu Iliyopangwa. 28 00:01:19,690 --> 00:01:22,450 Lakini tunahitaji kukimbia kupitia orodha tena ya kupata hiyo kabisa Iliyopangwa. 29 00:01:22,450 --> 00:01:29,250 Mbili ni chini ya 3, hivyo hatuna byta yao. 30 00:01:29,250 --> 00:01:31,700 >> Tatu ni chini ya 6, hivyo hatuna byta yao. 31 00:01:31,700 --> 00:01:35,500 Sita, ni mkubwa kuliko 5. Sisi swapped. 32 00:01:35,500 --> 00:01:38,460 Sita ni chini ya 9. Hatuna wabadilishane. 33 00:01:38,460 --> 00:01:42,170 Baada ya kupita pili kupitia, inaonekana kama hii: [2, 3, 5, 6, 9]. Perfect. 34 00:01:42,170 --> 00:01:44,680 Sasa, hebu andika katika pseudocode. 35 00:01:44,680 --> 00:01:48,450 Kimsingi, kwa kila kipengele katika orodha, tunahitaji kuangalia ni 36 00:01:48,450 --> 00:01:50,060 na kipengele moja kwa moja na haki yake. 37 00:01:50,060 --> 00:01:53,420 Kama wao ni nje ya utaratibu jamaa na kila mmoja - yaani kama kipengele juu ya kushoto 38 00:01:53,420 --> 00:01:56,810 ni mkubwa kuliko mmoja kulia - tunapaswa wabadilishane mambo mawili. 39 00:01:56,810 --> 00:02:01,270 >> Sisi kufanya hili kwa kila kipengele cha orodha, na tumekuwa alifanya moja kupita. 40 00:02:01,270 --> 00:02:05,160 Sasa sisi tu kufanya kupitisha-kupitia mara ya kutosha kuhakikisha orodha 41 00:02:05,160 --> 00:02:06,480 kikamilifu, ipasavyo Iliyopangwa. 42 00:02:06,480 --> 00:02:08,889 Lakini mara ngapi Je, tuna kupita katika orodha ya 43 00:02:08,889 --> 00:02:10,400 kuthibitisha kwamba sisi ni kosa? 44 00:02:10,400 --> 00:02:14,730 Naam, hali mbaya zaidi kesi ni kama tuna orodha kabisa nyuma. 45 00:02:14,730 --> 00:02:17,840 Kisha inachukua idadi ya kupita throughs-sawa na idadi 46 00:02:17,840 --> 00:02:19,730 ya mambo ya n-1. 47 00:02:19,730 --> 00:02:24,720 Kama hii haina mantiki intuitively, kufikiri ya kesi rahisi - orodha [2, 1]. 48 00:02:24,720 --> 00:02:28,430 >> Hii ni kwenda kuchukua kupitisha moja-kwa njia ya kutatua usahihi. 49 00:02:28,430 --> 00:02:33,060 [3, 2, 1] - kesi mbaya zaidi ni kwamba pamoja na mambo 3 sorted nyuma, 50 00:02:33,060 --> 00:02:34,830 ni kwenda kuchukua iterations 2 kwa aina. 51 00:02:34,830 --> 00:02:37,980 Baada iteration moja, ni [2, 1, 3]. 52 00:02:37,980 --> 00:02:39,550 mavuno ya pili safu sorted [1, 2, 3]. 53 00:02:39,550 --> 00:02:43,350 Hivyo unajua kamwe kwenda kupitia safu, kwa ujumla, 54 00:02:43,350 --> 00:02:46,790 zaidi ya mara n-1, ambapo n ni idadi ya vipengele katika safu. 55 00:02:47,090 --> 00:02:50,470 Ni wito Bubble Panga kwa sababu vipengele kubwa huwa na 'Bubble-up' 56 00:02:50,470 --> 00:02:51,950 kwa haki pretty haraka. 57 00:02:51,950 --> 00:02:53,980 Kwa kweli, algorithm hii ina kuvutia sana tabia. 58 00:02:53,980 --> 00:02:57,410 >> Baada iterations m kupitia safu nzima, 59 00:02:57,410 --> 00:02:59,000 rightmost vipengele m uhakika 60 00:02:59,000 --> 00:03:01,000 sorted katika nafasi zao sahihi. 61 00:03:01,000 --> 00:03:02,280 Kama unataka hii kuona mwenyewe, 62 00:03:02,280 --> 00:03:05,500 tunaweza kujaribu katika orodha ya nyuma kabisa [9, 6, 5, 3, 2]. 63 00:03:05,500 --> 00:03:08,220 Baada ya kupitisha moja kwa njia ya orodha nzima, 64 00:03:08,220 --> 00:03:09,220 [Sauti ya kuandika] 65 00:03:09,220 --> 00:03:18,790 [6, 9, 5, 3, 2], [6, 5, 9, 3, 2], [6, 5, 3, 9, 2], [6, 5, 3, 2, 9] 66 00:03:18,790 --> 00:03:21,250 kipengele rightmost 9 ni katika nafasi yake sahihi. 67 00:03:21,250 --> 00:03:24,760 Baada ya pili kupitisha-kupitia, 6 itakuwa na 'bubbled-up' kwa 68 00:03:24,760 --> 00:03:26,220 pili rightmost mahali. 69 00:03:26,220 --> 00:03:28,840 mambo mawili juu ya haki - 6 na 9 - itakuwa katika maeneo yao sahihi 70 00:03:28,840 --> 00:03:30,580 baada ya kwanza mbili kupita-throughs. 71 00:03:30,580 --> 00:03:32,590 >> Hivyo, jinsi gani tunaweza kutumia hii optimize algorithm? 72 00:03:32,590 --> 00:03:34,850 Naam, baada ya mmoja iteration kupitia safu 73 00:03:34,850 --> 00:03:37,690 sisi si kweli haja ya kuangalia kipengele rightmost 74 00:03:37,690 --> 00:03:39,200 kwa sababu tunajua ni Iliyopangwa. 75 00:03:39,200 --> 00:03:43,050 Baada iterations mbili, tunaweza kujua kwa uhakika rightmost mawili muhimu ni katika mahali. 76 00:03:43,050 --> 00:03:48,260 Hivyo, kwa ujumla, baada ya k iterations kupitia safu kamili, 77 00:03:48,260 --> 00:03:51,550 kuangalia vipengele k mwisho ni redundant tangu tunajua 78 00:03:51,550 --> 00:03:52,360 wao uko katika eneo sahihi tayari. 79 00:03:52,360 --> 00:03:54,870 >> Hivyo kama wewe ni kuchagua safu ya vipengele n, 80 00:03:54,870 --> 00:03:57,870 juu ya iteration kwanza - you'll kuwa na kuchambua wote wa mambo - kwanza n-0. 81 00:03:57,870 --> 00:04:04,170 On iteration pili, itabidi kuangalia wote wa mambo lakini mwisho - 82 00:04:04,170 --> 00:04:07,090 kwanza n-1. 83 00:04:07,090 --> 00:04:10,520 Optimization mwingine inaweza kuwa kwa kuangalia kama orodha tayari sorted 84 00:04:10,520 --> 00:04:11,710 baada ya kila iteration. 85 00:04:11,710 --> 00:04:13,900 Kama ni tayari Iliyopangwa, hatuna haja ya kufanya iterations yoyote zaidi 86 00:04:13,900 --> 00:04:15,310 kupitia orodha. 87 00:04:15,310 --> 00:04:16,220 Tunawezaje kufanya hivyo? 88 00:04:16,220 --> 00:04:19,360 Naam, kama sisi hatuwezi kufanya swaps yoyote juu ya kupitisha-kwa njia ya orodha, 89 00:04:19,360 --> 00:04:22,350 ni wazi kuwa orodha tayari alikuwa Iliyopangwa kwa sababu hatukuwa wabadilishane chochote. 90 00:04:22,350 --> 00:04:24,160 Hivyo sisi dhahiri hawana kutatua tena. 91 00:04:24,160 --> 00:04:27,960 >> Pengine unaweza initialize variable bendera iitwayo 'si vyema' kwa 92 00:04:27,960 --> 00:04:30,990 uongo na kubadilisha kwa kweli kama una wabadilishane vipengele yoyote juu ya 93 00:04:30,990 --> 00:04:32,290 moja iteration kupitia safu. 94 00:04:32,290 --> 00:04:35,350 Au vile vile, kufanya kinyume na uhesabu swaps kufanya 95 00:04:35,350 --> 00:04:37,040 yoyote iteration aliyopewa. 96 00:04:37,040 --> 00:04:40,040 Wakati wa mwisho wa iteration, ikiwa hakuwa wabadilishane yoyote ya vipengele, 97 00:04:40,040 --> 00:04:41,780 unajua orodha tayari sorted na wewe ni kosa. 98 00:04:41,780 --> 00:04:44,090 Bubble Panga, kama algorithms nyingine kuchagua, unaweza kuwa 99 00:04:44,090 --> 00:04:46,960 tweaked kufanya kazi kwa ajili ya mambo yoyote ambayo yana njia kuagiza. 100 00:04:46,960 --> 00:04:50,610 >> Hiyo ni, kutokana na mambo mawili muhimu una njia kusema kama moja ya kwanza 101 00:04:50,610 --> 00:04:53,770 ni mkubwa kuliko, sawa au chini ya pili. 102 00:04:53,770 --> 00:04:56,870 Kwa mfano, unaweza kutatua wa alfabeti kwa kusema 103 00:04:56,870 --> 00:05:00,520 kwamba 00:05:03,830 Au unaweza kutatua siku ya wiki ambapo Jumapili ni chini ya Jumatatu 105 00:05:03,830 --> 00:05:05,110 ambayo ni chini ya Jumanne. 106 00:05:05,110 --> 00:05:09,630 >> Bubble Panga ni kwa maana hakuna ufanisi sana au kufunga kuchagua algorithm. 107 00:05:09,630 --> 00:05:12,370 Mbaya wake-kesi Runtime ni Kubwa O ya n ² 108 00:05:12,370 --> 00:05:14,810 kwa sababu una kufanya iterations n kupitia orodha 109 00:05:14,810 --> 00:05:18,430 kuangalia vipengele vyote n kila kupitisha-kupitia, nxn = n ². 110 00:05:18,430 --> 00:05:22,730 Hii wakati kukimbia ina maana kwamba kama idadi ya vipengele wewe ni kuchagua kuongezeka, 111 00:05:22,730 --> 00:05:24,330 wakati kukimbia huongeza quadratically. 112 00:05:24,330 --> 00:05:27,330 >> Lakini kama ufanisi si tatizo kubwa ya mpango wako 113 00:05:27,330 --> 00:05:29,550 au kama wewe ni tu kuchagua idadi ndogo ya vipengele, 114 00:05:29,550 --> 00:05:31,660 unaweza kupata Bubble Panga muhimu kwa sababu 115 00:05:31,660 --> 00:05:33,360 ni moja ya algorithms rahisi kuchagua kuelewa 116 00:05:33,360 --> 00:05:34,250 na kwa kificho. 117 00:05:34,250 --> 00:05:37,270 Ni pia ni njia kubwa ya kupata uzoefu na kutafsiri nadharia 118 00:05:37,270 --> 00:05:40,220 algorithm katika kanuni halisi inayofanya kazi. 119 00:05:40,220 --> 00:05:43,000 Naam, hiyo ni Bubble Panga kwa ajili yenu. Shukrani kwa ajili ya kuangalia. 120 00:05:43,000 --> 00:05:44,000 CS50.TV