1、L,int e) int i,k=0; int *p,*q; p=L.elem; q=L.elem+L.length-1; for(i=1;i=L.length;i+) if(ep;q-) *q=*(q-1); *p=e; k=1; break; else p+; if(k=0) *(q+1)=e; L.length+;void main() void ListCreate(SqList &L,int i); int InitList_Sq(SqList &L); int SortInsert(SqList &L,int e); int i; SqList La; InitList_Sq(La
2、); ListCreate(La,3); int a; printf(需要插入的数字: ); scanf(a); SortInsert(La,a);插入后的元素列表变为:n for(i=0;La.length;%d ,La.elemi);/*void SortDeleSame(SqList & int *i; int *p; int *q; while(pnext=NULL; tail=L;n; s=(LinkList)malloc(sizeof(LNode);s-data); tail-next=s; tail=tail-next; tail-void Print_LinkList(Link
3、List & LNode* p; p=L- for(int i=0;,p- p=p-void ListInsert_LinkList(LinkList &L,int x) int k=0; LNode *pre,*p,*s; pre=L; while(pre&p) if(xdata) s-next=pre- pre- s-data=x; pre=pre- s=(LinkList)malloc(sizeof(LNode); pre-习题三、设计一算法,逆置带头结点的动态单链表L。/La为已知单链表,Lb为新建的一个单链表void Trunhead_LinkList(LinkList &La,Li
4、nkList &Lb,int n) LinkList p=La- LinkList s; Lb=(LinkList)malloc(sizeof(LNode); Lb-data=p-data;next=Lb-一元多项式之和实验#include typedef struct polynode float coef; /系数 int expn; /指数 struct polynode *next;polynode,*polylist;void create(polylist &L) /创建链表 int m; polylist p;请输入一元多项式项数:m); L=(polylist)malloc(s
5、izeof(polynode); p=L; for(int i=1;=m;i+) /利用循环,依次输入系数和指数 p-next=(polylist)malloc(sizeof(polynode);请输入第%d项的系数和指数:,i);%f %dp-coef,&expn);void display(polylist L) /显示链表内容%.0fx(%d)coef,p- while(p) if(p-coefnext ; pb=Lb- pc=(polylist)malloc(sizeof(polynode); Lc=pc; while (pa&pb) if(pa-expn=pb-expn) x=pa-
6、coef+pb-coef; if(x!=0) pc- pc=pc-coef=x;expn=pa-expn; pa=pa- pb=pb- else / 无同类项可合并,指数小者复制到C表 pc- pc=pc- if(pa-expn expn) /a的指数小于b的指数coef=pa- pc- pa=pa- else coef=pb-expn=pb- pb=pb- while(pa) /还剩下a多项式 pc- pc- while(pb) /还剩下b多项式 pb=pb-void subtract(polylist La,polylist Lb,polylist &Lc) /减法函数 polylist
7、pa,pb,pc; while(pa& if(pa- x=pa-coef-pb- pc- else /无同类项可合并,指数小者复制到C表 if (pa-expncoef=-pb- while (pb) /还剩下b多项式void main() /主函数 polylist La,Lb,Lc,Ld; create(La); create(Lb);一元多项式1: display(La);一元多项式2: display(Lb); add(La,Lb,Lc);加的结果: display(Lc); subtract(La,Lb,Ld);减的结果 display(Ld);哈夫曼树实验string.h char
8、 ch; int weight; int parent,lchild,rchild;HTNode,*HuffmanTree;typedef char*HuffmanCode;void CreateHuffmanTree(HuffmanTree &HT,int w,char ch,int n);void Select(HuffmanTree HT,int n,int &s1,int &s2);void HTCoding(HuffmanTree HT,HuffmanCode &HC,int n);void PrintCode(HuffmanCode HC,int n,char ch);double
9、 AverageLength(HuffmanTree HT,HuffmanCode HC,int n);void DeCode(HuffmanTree HT,int n); int n; char arrch20; int arrweight20; double avlength; HuffmanTree HT; HuffmanCode HC;请输入要输入字母的个数:n); while(ch=getchar()!=n if(n20|n2) exit(0);请输入第%d个字母:,i+1);%carrchi);请输入该字母权重:arrweighti); CreateHuffmanTree(HT,a
10、rrweight,arrch,n); HTCoding(HT,HC,n); PrintCode(HC,n,arrch); avlength=AverageLength(HT,HC,n);平均编码长度为:%.2fn,avlength);请输入要解码的数据: DeCode(HT,n); free(HCi); free(HC); free(HT);HT,int w,char ch,int n) int i,m,s1,s2; m=2*n-1; HT=(HuffmanTree)malloc(m+1)*sizeof(HTNode);=n; HTi.weight=wi-1; HTi.parent=0; HT
11、i.lchild=0; HTi.rchild=0; HTi.ch=chi-1; for(i=n+1; HTi.weight=0; HTi.ch=0; Select(HT,i-1,s1,s2); HTs1.parent=i; HTs2.parent=i; HTi.lchild=s1; HTi.rchild=s2; HTi.weight=HTs1.weight+HTs2.weight;s2) int min; min=1000; if(HTi.parent=0&HTi.weight=min) min=HTi.weight; s1=i;=min&i!=s1) s2=i;HC,int n) int i
12、,j,k,start; int f; int c; char* cd; HC=(HuffmanCode)malloc(n)*sizeof(char*); cd=(char*)malloc(n*sizeof(char); cdn-1=+i) start=n-1; for(c=i,f=HTi.parent;f!=0;c=f,f=HTf.parent) if(HTf.lchild=c) cd-start=0 else1 HCi-1=(char*)malloc(n-start)*sizeof(char); for(j=start,k=0;j+,k+) HCi-1k=cdj; free(cd);void
13、 PrintCode(HuffmanCode HC,int n,char ch)%c的编码是%sn,chi,HCi);double AverageLength(HuffmanTree HT,HuffmanCode HC,int n) int i,j; int s1=0,s2=0; s1=s1+HTi.weight; for(i=0,j=1;i+,j+) s2=s2+HTj.weight*strlen(HCi); return s2*1.0/s1;void DeCode(HuffmanTree HT,int n) char endflag=# i=2*n-1;ch); while(ch!=endflag) if(ch=) i=HTi.lchild; else i=HTi.rchild; if(HTi.lchild=0) printf(,HTi.ch); i=2*n-1; if(HTi.lchild!=0)&(i!=2*n-1) printf(n未能完全解码n求最小生成树算法实验#in
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