数据结构作业(3

1、班级：2011级计算机科学与技术一班姓名：弥沛学号:2011222223日期：2012年9月28日题目一：整理两个一元多项式相加程序代码:stdio.hstdiib.h struct node#i nclude#i ncludetyp edefint cof,exp;struct node * next; no de;void creat( struct node *h) int x,y;struct node *r, *p;r=h;printK input x,y:);scanf( %d,%d, &x, &y); while (x !=- 999)p=( struct node*)mallo

2、c( sizeof (struct no de); p- cof =x;p-ex p=y;r-n ext =p;r=p;scanf( %d,%d, &x, &y);r- next =NULL;void outline( struct node *h)struct node *p;p=h- n ext;while (p != NULL)void add (node *h1,node *h2) int x;struct node *r, * p, *q,*s;p=h1-n ext;q=h2- n ext;r=h1;while (p !=NUL&|!=NULL)ifrrP(P - expq- exp

3、)- next =p;=P;=p-n ext;printf( %dxA%d-,p - cof,p -e xp);p=p-n ext;printf( n); elserr q elsex=p-cof +q- cof;if (x =0)/s=p;p=p- n ext;/free(s);/s=q;q=q- n ext;/free(s);elsep- cof =x; r-n ext =p; r=p;if (p - expvq- exp )- next =q;=q;=q-n ext;ififp=Fh n ext;/s=q；q=q- n ext;/free(s);(p != NULL)r-n ext =p

4、;(q!=NULL)r-n ext =q;free(h2);运行结果：void main()struct node *h1,*h2;h1 =( structnode*)malloc( sizeof (structn ode);h2 =( struct node*)malloc( sizeof (struct no de);creat(h1);outli ne(h1);creat(h2);outli ne(h2);add(h1,h2);outli ne(h1); CrJM SOFTC VuYdfiXbinwwtem p. exeinput x y： 5, 4 6, 32 4, 1 L, 0-99

5、95s*4-6?:*3-5s*2-4x*l-lx*0-input X, y： 3, 3 5, 2 1, 0-9995x*3-5x*2-lx*0-ox - 4-11k3-10x2-4k1-2x O-Press any key to continue题目将两个有序单链表合并，并保持它的有序性题目二：单链表中数据的就地逆置。1、就地逆置代码:mai n()#include #include typedef struct Nodeint data;struct Node * n ext; Node, *LinkList;void create( struct Node *head)struct Nod

6、e *p, *q;int flag =1,x;p =head;printf( input x:);while (flag)q=( struct Node*)malloc( sizeof (struct Node); scanf( %d, &x);if (x !=- 999)q-data =x;p -n ext =q;p =p- n ext;elseflag =0;p -n ext =NULL; void NIZHI(Li nkList L) Node *p, *q; p =L- n ext;L- next =NULL; while (p !=NULL) 运行结果：q=p-n ext;p- n

7、ext =L- n ext;L- n ext =p;p=q;voidLin kList l;Node *p;l =(Node* )malloc( sizeof (Node);l - next = NULL;printf(请输入链表数据,以-999 结束!n);create(l);printf(输入的单链表为:n); p = I -next;while (p !=NULL)printf( %dn, p-data); p=p-n ext; NIZHI(l);printf(逆置后的单链表为:n);p = I -next;while (p !=NULL)printf( %dn,p-data); p=p

8、-n ext;scanf(- data =x;- n ext =NULL;- n ext =p;=p;%d, &x);node * p= head-next; n);(p !=NULL)#include #include struct nodeint data;struct node *next;void create( struct node * head)/*尾插法创建链表*/struct node *p, *r=head; int x;scanf(%d, &x);while (x !=- 999)p =( struct node*)malloc( sizeof (struct no de

9、);PPrrvoid outline(struct node *head)/*链表的输出*/structprintf(whileprintf( %5d, p-data);p =p- n ext;void mergelist( struct node *head1, struct node *head2) /*将两个有序链表合并*/struct node *p, *q, *r;=head1- n ext;=head2- n ext;=head1;while (p != NUL&|!= NULL)-n ext =p;=P；=p- next;-n ext =q;=q;=q- next;if (p- data v=q-data) rrP elserrqif (q =NULL)r -n ext = p; elser -n ext =q;mai n()struct node *head1, *head2; headl =( structnode*)malloc( sizeof (struct node);head2 =( structnode*)malloc( sizeof (struct node);head1 -n ext=NULL;head2 - next =NULL; create(head1);create(head2);printsout put arran ge