電大數(shù)據(jù)結(jié)構(gòu)填空題小抄參考

1、專業(yè)好文檔1把數(shù)據(jù)存儲(chǔ)到計(jì)算機(jī)中,并具體體現(xiàn)數(shù)據(jù)之間的邏輯結(jié)構(gòu)稱為 物理（存儲(chǔ)）結(jié)構(gòu)。2設(shè)有一個(gè)不帶頭結(jié)點(diǎn)的單向循環(huán)鏈表，結(jié)點(diǎn)的指針域?yàn)閚ext，指針p指向尾結(jié)點(diǎn)，現(xiàn)要使p指向第一個(gè)結(jié)點(diǎn)，可用語句 p=p-next 。3結(jié)構(gòu)中的數(shù)據(jù)元素存在一對(duì)一的關(guān)系稱為 線性 結(jié)構(gòu)。4要在一個(gè)帶頭結(jié)點(diǎn)的單向循環(huán)鏈表中刪除頭結(jié)點(diǎn)，得到一個(gè)新的不帶頭結(jié)點(diǎn)的單向循環(huán)鏈表，若結(jié)點(diǎn)的指針域?yàn)閚ext，頭指針為head，尾指針為p，則可執(zhí)行head=head- next; p-next=head。5在雙向鏈表中，每個(gè)結(jié)點(diǎn)有兩個(gè)指針域，一個(gè)指向 結(jié)點(diǎn)的直接后繼，另一個(gè)指向 結(jié)點(diǎn)的直接前驅(qū)。6設(shè)有一個(gè)非空的鏈棧，棧頂指針為

2、hs，要進(jìn)行出棧操作，用x保存出棧結(jié)點(diǎn)的值，棧結(jié)點(diǎn)的指針域?yàn)閚ext，數(shù)據(jù)域?yàn)閐ata，則可執(zhí)行x= hs-data；和hs= hs-next；7設(shè)有一個(gè)頭指針為head的單向鏈表，p指向表中某一個(gè)結(jié)點(diǎn)，且有p-next= =null,通過操作p-next=head ，就可使該單向鏈表構(gòu)造成單向循環(huán)鏈表。8循環(huán)隊(duì)列的最大存儲(chǔ)空間為maxsize，隊(duì)頭指針為f，隊(duì)尾指針為r，當(dāng)（r+1）%maxsize=f 時(shí)表明隊(duì)列已滿。9從一個(gè)棧頂指針為h的鏈棧中刪除一個(gè)結(jié)點(diǎn)時(shí)，用x保存被刪結(jié)點(diǎn)的值，可執(zhí)行x=h-data;和h=h-next。(結(jié)點(diǎn)的指針域?yàn)閚ext)10程序段 int count=0;

3、char *s=” abcd”; while(*s!=0)s+;count+; 執(zhí)行后count= 411兩個(gè)串相等的充分必要條件是 串長度相等且對(duì)應(yīng)位置的字符相等 。12一棵二叉樹總結(jié)點(diǎn)數(shù)為11，葉結(jié)點(diǎn)數(shù)為5，該樹有 4 個(gè)雙分支結(jié)點(diǎn)，2個(gè)單分支結(jié)點(diǎn)。13對(duì)二叉樹的遍歷可分為 先序、中序、后序、層次 四種不同的遍歷次序。14設(shè)一棵完全二叉樹，其最高層上最右邊的葉結(jié)點(diǎn)的編號(hào)為偶數(shù)，該葉節(jié)點(diǎn)的雙親結(jié)點(diǎn)的編號(hào)為9，該完全二叉樹一共有 18 個(gè)結(jié)點(diǎn)。15一棵有n個(gè)葉結(jié)點(diǎn)的二叉樹，其每一個(gè)非葉結(jié)點(diǎn)的度數(shù)都為2，則該樹共有 2n-1 個(gè)結(jié)點(diǎn)。16雙向循環(huán)鏈表中，p指向表中某結(jié)點(diǎn)，則通過p可以訪問到p所指

4、結(jié)點(diǎn)的直接后繼結(jié)點(diǎn)和直接前驅(qū)結(jié)點(diǎn)，這種說法是 正確 的。17一棵有14個(gè)結(jié)點(diǎn)的完全二叉樹，則它的最高層上有 7 個(gè)結(jié)點(diǎn)。18棧和隊(duì)列的操作特點(diǎn)分別是 先進(jìn)后出 和 先進(jìn)先出。19如圖2所示的二叉樹，其先序遍歷序列為 abdgcefhi。efgibachd 20折半查找只適用于 順序存儲(chǔ)結(jié)構(gòu) 存儲(chǔ)的有序表 。21哈希函數(shù)是記錄關(guān)鍵字值與該記錄 存儲(chǔ)地址 之間所構(gòu)造的對(duì)應(yīng)關(guān)系。22深度為k的二叉樹最多有 2k-1 結(jié)點(diǎn)。23二叉樹排序中任一棵子樹都是二叉排序樹，這種說法是 正確 的。24串的兩種最基本的存儲(chǔ)方式是 順序存儲(chǔ) 和 鏈?zhǔn)酱鎯?chǔ)。1通常數(shù)據(jù)的邏輯結(jié)構(gòu)包括 集合；線性；樹形；圖狀 四種類型。

5、2結(jié)構(gòu)中的元素之間存在多對(duì)多的關(guān)系稱為 圖狀 結(jié)構(gòu)。3設(shè)有一個(gè)單向鏈表，結(jié)點(diǎn)的指針域?yàn)閚ext，頭指針為head，p指向尾結(jié)點(diǎn)，為了使該單向鏈表改為單向循環(huán)鏈表，可用語句 p-next=head 。4設(shè)有一個(gè)單向循環(huán)鏈表，結(jié)點(diǎn)的指針域?yàn)閚ext，頭指針為head，指針p指向表中某結(jié)點(diǎn)，若邏輯表達(dá)式 p-next= =head的結(jié)果為真，則p所指結(jié)點(diǎn)為尾結(jié)點(diǎn)。5設(shè)有一個(gè)單向循環(huán)鏈表，頭指針為head，鏈表中結(jié)點(diǎn)的指針域?yàn)閚ext，p指向尾結(jié)點(diǎn)的直接前驅(qū)結(jié)點(diǎn)，若要?jiǎng)h除尾結(jié)點(diǎn)，得到一個(gè)新的單向循環(huán)鏈表，可執(zhí)行操作p-next=head 。 6設(shè)有一個(gè)鏈棧，棧頂指針為hs，現(xiàn)有一個(gè)s所指向的結(jié)點(diǎn)要入棧

6、，則可執(zhí)行操作s- next=hs; hs=s 。7在一個(gè)鏈隊(duì)中，f和r分別為隊(duì)頭和隊(duì)尾指針，隊(duì)結(jié)點(diǎn)的指針域?yàn)閚ext，則插入一個(gè)s所指結(jié)點(diǎn)的操作為r-next=s；r=s；8在一個(gè)鏈隊(duì)中，f和r分別為隊(duì)頭和隊(duì)尾指針，隊(duì)結(jié)點(diǎn)的指針域?yàn)閚ext，s指向一個(gè)要入隊(duì)的結(jié)點(diǎn)，則入隊(duì)操作為r-next=s；r=s；9循環(huán)隊(duì)列的隊(duì)頭指針為f，隊(duì)尾指針為r，當(dāng) r= =f 時(shí)表明隊(duì)列為空。 10循環(huán)隊(duì)列的最大存儲(chǔ)空間為maxsize=6，采用少用一個(gè)元素空間以有效地判斷棧空或棧滿，若隊(duì)頭指針front=4，當(dāng)隊(duì)尾指針rear= 3 時(shí)隊(duì)滿，隊(duì)列中共有5個(gè)元素。11a在存儲(chǔ)時(shí)占 1個(gè)字節(jié)。“a”在存儲(chǔ)時(shí)占 2

7、 個(gè)字節(jié)。12程序段 char *s=”abcd”;n=0; while(*s!=0) if(*sa&*snext= p-next和p-next=s;的操作。4要在一個(gè)單向鏈表中刪除p所指向的結(jié)點(diǎn)，已知q指向p所指結(jié)點(diǎn)的直接前驅(qū)結(jié)點(diǎn)，若鏈表中結(jié)點(diǎn)的指針域?yàn)閚ext，則可執(zhí)行q-next= p-next。5設(shè)有一個(gè)非空的鏈棧，棧頂指針為hs，要進(jìn)行出棧操作，用x保存出棧結(jié)點(diǎn)的值，棧結(jié)點(diǎn)的指針域?yàn)閚ext，則可執(zhí)行x=hs-data;hs=hs-next。6設(shè)有一個(gè)鏈棧，棧頂指針為hs，現(xiàn)有一個(gè)s所指向的結(jié)點(diǎn)要入棧，則可執(zhí)行操作s-next=hs和hs=s；7在一個(gè)不帶頭結(jié)點(diǎn)的非空鏈隊(duì)中，f和r分

8、別為隊(duì)頭和隊(duì)尾指針，隊(duì)結(jié)點(diǎn)的數(shù)據(jù)域?yàn)閐ata，指針域?yàn)閚ext，若要進(jìn)行出隊(duì)操作，并用變量x存放出隊(duì)元素的數(shù)據(jù)值，則相關(guān)操作為x=f-data; f=f-next。8在一個(gè)不帶頭結(jié)點(diǎn)的非空鏈隊(duì)中，f和r分別為隊(duì)頭和隊(duì)尾指針，隊(duì)結(jié)點(diǎn)的數(shù)據(jù)域?yàn)閐ata，指針域?yàn)閚ext，若要進(jìn)行出隊(duì)操作，并用變量x存放出隊(duì)元素的數(shù)據(jù)值，則相關(guān)操作為 x=f-data; f=f-next;。9循環(huán)隊(duì)列的最大存儲(chǔ)空間為maxsize=8，采用少用一個(gè)元素空間以有效的判斷棧空或棧滿，若隊(duì)頭指針front=4，則當(dāng)隊(duì)尾指針rear= 4時(shí)，隊(duì)列為空，當(dāng)rear= 2時(shí)，隊(duì)列有6個(gè)元素。10，順序存儲(chǔ)字符串“abcd”需要

9、占用 5 個(gè)字節(jié)。11稀疏矩陣存儲(chǔ)時(shí)，采用一個(gè)由 行號(hào)；列號(hào)；非零元 3部分信息組成的三元組唯一確定矩陣中的一個(gè)非零元素。12一棵二叉樹葉結(jié)點(diǎn)（終端結(jié)點(diǎn)）數(shù)為5，單分支結(jié)點(diǎn)數(shù)為2，該樹共有11個(gè)結(jié)點(diǎn)。13一棵二叉樹順序編號(hào)為6的結(jié)點(diǎn)（樹中各結(jié)點(diǎn)的編號(hào)與等深度的完全二叉中對(duì)應(yīng)位置上結(jié)點(diǎn)的編號(hào)相同），若它存在右孩子，則右孩子的編號(hào)為 13。14設(shè)一棵完全二叉樹，其最高層上最右邊的葉結(jié)點(diǎn)的編號(hào)為奇數(shù)，該葉節(jié)點(diǎn)的雙親結(jié)點(diǎn)的編號(hào)為10，該完全二叉樹一共有 21 個(gè)結(jié)點(diǎn)。15結(jié)構(gòu)中的數(shù)據(jù)元素存在多對(duì)多的關(guān)系稱為 圖狀 結(jié)構(gòu)。16結(jié)構(gòu)中的數(shù)據(jù)元素存在一對(duì)多的關(guān)系稱為 樹形 結(jié)構(gòu)。17結(jié)構(gòu)中的數(shù)據(jù)元素存在一對(duì)

10、多的關(guān)系稱為 樹形 結(jié)構(gòu)。18結(jié)構(gòu)中的數(shù)據(jù)元素存在一對(duì)一的關(guān)系稱為 線性 結(jié)構(gòu)。19如圖2所示的二叉樹，其前序遍歷序列為abdefcg。gfabdec21在隊(duì)列的順序存儲(chǔ)結(jié)構(gòu)中，當(dāng)插入一個(gè)新的隊(duì)列元素時(shí), 尾 指針的值增1，當(dāng)刪除一個(gè)元素隊(duì)列時(shí)，頭 指針的值增1。22圖的深度優(yōu)先搜索和廣度優(yōu)先搜索序列不一定是唯一的。此斷言是 正確 的。(回答正確或不正確) 23循環(huán)隊(duì)列的引入，目的是為了克服 假上溢。24按某關(guān)鍵字對(duì)記錄序列排序，若關(guān)鍵字 相等 的記錄在排序前和排序后仍保持它們的前后關(guān)系，則排序算法是穩(wěn)定的，否則是不穩(wěn)定的。winger tuivasa-sheck, who scored tw

11、o tries in the kiwis 20-18 semi-final win over england, has been passed fit after a lower-leg injury, while slater has been named at full-back but is still recovering from a knee injury aggravated against usa.both sides boast 100% records heading into the encounter but australia have not conceded a

12、try since josh charnleys effort in their first pool match against england on the opening day.aussie winger jarryd hayne is the competitions top try scorer with nine, closely followed by tuivasa-sheck with eight.but it is recently named rugby league international federation player of the year sonny b

13、ill williams who has attracted the most interest in the tournament so far.the kiwi - with a tournament high 17 offloads - has the chance of becoming the first player to win the world cup in both rugby league and rugby union after triumphing with the all blacks in 2011.id give every award back in a h

14、eartbeat just to get across the line this weekend, said williams.the (lack of) air up there watch mcayman islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 150th anniversary celebrations and will attend citys premier league match at chelsea on su

15、nday.i am going to be at the match tomorrow and i have asked to meet yaya toure, he told bbc sport.for me its about how he felt and i would like to speak to him first to find out what his experience was.uefa hasopened disciplinary proceedings against cskafor the racist behaviour of their fans during

16、citys 2-1 win.michel platini, president of european footballs governing body, has also ordered an immediate investigation into the referees actions.cska said they were surprised and disappointed by toures complaint. in a statement the russian side added: we found no racist insults from fans of cska.

17、 baumgartner the disappointing news: mission aborted.the supersonic descent could happen as early as sunda.the weather plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover.

18、the balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 m

19、iles/9.17 kilometers) and into the stratosphere. as he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.the balloon will slowly drift to the edge of space at 120,000 feet ( then, i would assume, he will slowly step out onto something resembling an olympic diving pl

20、atform.they blew it in 2008 when they got caught cold in the final and they will not make the same mistake against the kiwis in manchester.five years ago they cruised through to the final and so far history has repeated itself here - the last try they conceded was scored by englands josh charnley in

21、 the opening game of the tournament.that could be classed as a weakness, a team under-cooked - but i have been impressed by the kangaroos focus in their games since then.they have been concentrating on the sort of stuff that wins you tough, even contests - strong defence, especially on their own goal-line, completing sets and a good kick-ch