数据结构题一道,请高手帮忙~具体算法是怎么算的! 一道数据结构的题,请帮忙改错。。在线等

\u6570\u636e\u7ed3\u6784\u9ad8\u624b\u6765\u5e2e\u5fd9\uff08\u7b80\u7b54\u9898\u3001\u7b97\u6cd5\u9898\uff09

1\u3001\u987a\u5e8f\u5b58\u50a8\u65b9\u5f0f\u53ea\u80fd\u7528\u4e8e\u5b58\u50a8\u7ebf\u6027\u7ed3\u6784\u3002\uff08 N \uff09
2\u3001\u6570\u7ec4\u4e0d\u9002\u5408\u4f5c\u4e3a\u4e8c\u53c9\u6811\u7684\u5b58\u50a8\u7ed3\u6784\u3002\uff08 N \uff09
3\u3001\u4e32\u662f\u4e00\u79cd\u6570\u636e\u5bf9\u8c61\u548c\u64cd\u4f5c\u90fd\u7279\u6b8a\u7684\u7ebf\u6027\u8868\u3002\uff08 Y \uff09
4\u3001\u7ebf\u6027\u8868\u91c7\u7528\u94fe\u8868\u5b58\u50a8\u65f6\uff0c\u7ed3\u70b9\u548c\u7ed3\u70b9\u5185\u90e8\u7684\u5b58\u50a8\u7a7a\u95f4\u53ef\u4ee5\u662f\u4e0d\u8fde\u7eed\u7684\u3002\uff08 Y \uff09
5\u3001\u6808\u548c\u961f\u5217\u90fd\u662f\u9650\u98de\u8fc7\u6d77\u82f1\u8bed\u89d2\u5236\u5b58\u53d6\u70b9\u7684\u7ebf\u6027\u7ed3\u6784\u3002\uff08 Y \uff09
6\u3001\u4e00\u4e2a\u5e7f\u4e49\u8868\u53ef\u4ee5\u4e3a\u5176\u5b83\u5e7f\u4e49\u8868\u6240\u5171\u4eab\u3002\uff08 N \uff09
7\u3001\u6811\u7684\u5ea6\u662f\u6307\u6811\u5185\u7ed3\u70b9\u7684\u5ea6\u3002\uff08 Y \uff09
8\u3001\u4e00\u68f5\u4e00\u822c\u6811\u7684\u7ed3\u70b9\u7684 \u6839\u6b21\u5e8f\u904d\u5386\u548c\u540e\u6839\u6b21\u5e8f\u904d\u5386\u5206\u522b\u4e0e\u5176\u76f8\u5e94\u4e8c\u53c9\u6811\u7684\u7ed3\u70b9\u524d\u5e8f\u904d\u5386\u4f46\u662f\u548c\u540e\u5e8f\u5c31\u51e0\u53f7\u56de\u5bb6\u8c6b\u5267\u904d\u5386\u662f\u4e00\u81f4\u7684\u3002\uff08 N \uff09
9\u3001\u65e0\u5411\u56fe\u7684\u90bb\u63a5\u77e9\u9635\u4e00\u5b9a\u5bf9\u79f0\u77e9\u9635\uff0c\u6709\u5411\u56fe\u7684\u90bb\u63a5\u77e9\u9635\u4e00\u5b9a\u662f\u975e\u5bf9\u79f0\u77e9\u9635\u3002\uff08 N \uff09
10\u3001\u6392\u5e8f\u7b97\u6cd5\u4e2d\u7684\u6bd4\u8f83\u6b21\u6570\u4e0e\u521d\u59cb\u5143\u7d20\u5e8f\u5217\u7684\u6392\u5217\u65e0\u5173\u3002\uff08 X \uff09

1\u3001 \u8bbe\u76ee\u6807\u4e32t=\u201cabaabcc\u201d\uff0c\u6a21\u5f0f\u4e32P=\u201caabc\u201d\uff0c\u8bd5\u63cf\u8ff0\u6839\u636e\u7a77\u4e3e\u6a21\u5f0f\u5339\u914d\u7b97\u6cd5\u8fdb\u884c\u5339\u914d\u7684\u8fc7\u7a0b\u3002

abaabcc
aabc d=0, fail
aabc d=1, fail
aabc d=2, success, return 2

2\u3001 \u8bbe\u7528\u4e8e\u901a\u8baf\u7684\u7535\u6587\u6cd5\u75318\u4e2a\u5b57\u6bcd\u7ec4\u6210\uff0c\u5b57\u6bcd\u5728\u7535\u6587\u4e2d\u51fa\u73b0\u7684\u9891\u7387\u5206\u522b\u4e3a7\uff0c9\uff0c2\uff0c6\uff0c32\uff0c3\uff0c21\uff0c10\uff0c\u8bd5\u4e3a\u8fd98\u4e2a\u5b57\u6bcd\u8bbe\u8ba1\u4e0d\u7b49\u957fHuffman\u7f16\u7801\uff0c\u5e76\u7ed9\u51fa\u8be5\u7535\u6587\u7684\u603b\u7801\u6570\uff08\u8981\u6c42\u753b\u51faHuffman\u7f16\u7801\u6570\uff09\u3002

\u5047\u8bbe\uff1a
a:7 b:9 c:2 d:6 e:32 f:3 g:21 h:10
\u6392\u5e8f\uff1a
(c:2) (f:3) (d:6) (a:7) (b:9) (h:10) (g:21) (e:32)
\u6309\u4f18\u5148\u7ea7\u5408\u5e76\uff1a
((c[0],f[1]):5) (d:6) (a:7) (b:9) (h:10) (g:21) (e:32)
(a:7) (b:9) (h:10) (((c[00],f[01]),d[1]):11) (g:21) (e:32)
(h:10) (((c[00],f[01]),d[1]):11) ((a[0],b[1]):16) (g:21) (e:32)
((a[0],b[1]):16) ((h[0],((c[100],f[101]),d[11])):21) (g:21) (e:32)
(g:21) (e:32) (((a[00],b[01]),(h[10],((c[1100],f[1101]),d[111]))):37)
(((a[00],b[01]),(h[10],((c[1100],f[1101]),d[111]))):37) ((g[0],e[1]):43)
((((a[000],b[001]),(h[010],((c[01100],f[01101]),d[0111]))),(g[10],e[11])):80)
a:000 3*7=21
b:001 3*9=37
c:01100 5*2=10
d:0111 4*6=24
e:11 2*32=64
f:01101 5*3=15
g:10 2*21=42
h:010 3*10=30

\u8fd9\u662f\u6309\u4f60\u7684\u601d\u8def\u6539\u7684\uff0c\u6ca1\u95ee\u9898\u4e86\u3002\u4f46\u6211\u89c9\u5f97\u4e0d\u592a\u5408\u9002\uff0c\u56e0\u4e3a\u4f60\u6ca1\u6709\u5efa\u7acb\u4e24\u4e2a\u94fe\u8868\uff0c
#include
#include
typedef struct node{

int data;

struct node *link;

}LNode,*LinkList;


LinkList INSERT(LinkList list,int item)
{

LinkList p,q,r;

q=(LinkList)malloc(sizeof(LNode));

q->data=item;

q->link=NULL;

p=list;

r=NULL;

while(p!=NULL)
{

if((p->data)==(q->data))

break;

else if((p->data)data))
{

r=p;

p=p->link;

}

else
{
if(r==NULL)
{
q->link=list;
list=q;
}
else
{
r->link=q;

q->link=p;
}


break;

}

}

if(p==NULL)
{

r->link=q;

q->link=NULL;

}

return list;

}


int main()
{

LinkList p,r,lista;

int a,i;

char c;

lista=NULL;

do
{

scanf("%d",&a);

c=getchar();

p=(LinkList)malloc(sizeof(LNode));

p->data=a;

p->link=NULL;

if(lista==NULL)

lista=p;

else

r->link=p;

r=p;

}
while(c==' ');

do
{

scanf("%d",&a);

c=getchar();

lista=INSERT(lista,a);

}
while(c==' ');

p=lista;

while(p!=NULL)
{

printf("%d ",p->data);


p=p->link;

}

return 0;

}

用查找每个元素的概率乘找到该元素需要遍历的长度后相加即得到结果。

avgLen = (1/24)*1 + (1/6)*2 + (2/3)*3 + (1/8)*4 = 23/8

故选A

1*(1/24)+2*(1/6)+3*(2/3)+4*(1/8)=23/8
即答案A

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