排序及查找方法
#include <malloc.h>
#include<stdio.h>
#define N 11
/*用监视哨查找*/
int search(int array[],int n,int k)
{int i;
i=n-1;
array[0]=k;
while(array[i]!=k) i--;
return(i);
}
/*折半查找法*/
int halfsearch(int array[],int n,int k)
{int i,j,mid;
i=1;j=n;
while(i<=j)
{mid=(i+j)/2;
if(k==array[mid]) return(mid);
else if(k<array[mid]) j=mid-1;
else i=mid+1;
}
return(0);
}
/*冒泡排序法*/
void mpsort(int array[])
{int i,j,a;
a=0;
for(i=1;i<N;i++)
for(j=i+1;j<N;j++)
if(array[i]>array[j])
{a=array[i];
array[i]=array[j];
array[j]=a;}
}
/*直接插入排序*/
void insertsort(int array[])
{int i,j;
for(i=2;i<N;i++)
{array[0]=array[i];
j=i-1;
while(array[0]<array[j])
{array[j+1]=array[j--];
array[j+1]=array[0];
}
}
}
/*建立*/
void creat(int array[])
{int i;
printf("enter the array:\n");
for(i=1;i<N;i++)
scanf("%d",&array[i]);
}
/*显示*/
void print(int array[])
{int i;
printf("The numbers after sort is:\n");
for(i=1;i<N;i++)
printf("%d ",array[i]);
printf("\n");
}
main()
{int a[11],i,x,chang;
/*printf("enter the array\n");
for(i=1;i<11;i++)
scanf("%d",&a[i]);*/
aga:
printf("\nchang:1: use watching method finding\n 2:use half method finding\n 3: use directness intsert method sort\n 4:use bubble up method sort\n 5:exit\n");
scanf("%d",&chang);
switch (chang)
{case 1:
{creat(a);
printf("Please int the search number:\n");
scanf("%d",&x);
printf("The number station is:%d\n",search(a,N,x));
goto aga;
}
case 2:
{ creat(a);
insertsort(a);
print(a);
printf("Please int the search number:\n");
scanf("%d",&x);
printf("The number station is:%d\n",halfsearch(a,N,x));
goto aga;
}
case 3:
{creat(a);
insertsort(a);
print(a);
goto aga;
}
case 4:
{creat(a);
mpsort(a);
print(a);
goto aga;
}
case 5:{ printf("exit!\n");break;}
default:{printf("Error!\n"); goto aga;}
}
}
二、线性链表的存储实现
struct LNODE{
ElemType data;
struct LNODE *next;
};
typedef struct LNODE LNode;
typedef struct LNODE * LinkList;
1初始化操作
Status Init_L(LinkList L){
if (L=(LinkList *)malloc(sizeof(LNode)))
{L->next=NULL;return 1;}
else return 0;
}
2插入操作
Status ListInsert_L(LinkList &L,int i,ElemType e){
p=L,j=0;
while(p&&j<i-1){p=p->next;++j;}
if(!p||j>i-1) return ERROR;
s=(LinkList)malloc(sizeof(LNode));
s->data=e;s->next=p->next;
p->next=s;
return OK;
}//ListInsert_L
3删除操作
Status ListDelete_L(LinkList &L,int i,ElemType &e){
p=L,j=0;
while(p&&j<i-1){p=p->next;++j;}
if(!p->next||j>i-1) return ERROR;
q=p->next;p->next=q->next;
e=q->data;free(q);
return OK;
}//ListDelete_L
4取某序号元素的操作
Status GetElem_L(LinkList &L,int i,ElemType &e){
p=L->next,j=1;
while(p&&j<i){p=p->next;++j;}
if(!p||j>i) return ERROR;
e=p->data;
return OK;
}//GetElem_L
5归并两个单链表的算法
void MergeList_L(LinkList &La,LinkList &Lb,LinkList &Lc){
//已知单链线性表La和Lb的元素按值非递减排列
//归并后得到新的单链线性表Lc,元素也按值非递减排列
pa=La->next;pb=Lb->next;
Lc=pc=La;
while(pa&&pb){
if(pa->data<=pb->data){
pc->next=pa;pc=pa;pa=pa->next;
}else{pc->next=pb;pc=pb;pb=pb->next;}
}
pc->next=pa?pa:pb;
free(Lb);
}//MergeList_L
头指针与头结点的区别:
头指针只相当于结点的指针域,头结点即整个线性链表的第一个结点,它的数据域可以放数据元素,也可以放 线性表的长度等附加信息,也可以不存储任何信息。
典型的约瑟夫环问题。原问题比你的问题要复杂一点。我以前写的程序:
1.用数组。
# include "stdio.h"
# define SIZE 100
main()
{
int m,n,i;
int array[SIZE];
printf("约瑟夫环求解,当前设置最大人数为%d.\n",SIZE);
printf("报数上限:\n");
scanf("%d",&m);
printf("总人数为:\n");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("第%d人的密码为:",i+1);
scanf("%d",&array);
}
joseph(array,m,n);
}
int joseph(a,m,n)
int a[],m,n;
{
int b[SIZE]; /*计录编号数组.*/
int i; /*计数器.*/
int flag=0;
int code; /*删取人的号码.*/
int sum=n; /*现存人数.*/
int point=0; /*当前报数人的位置.*/
int num=m;
for(i=0;i<n;i++) /*计录数组.*/
{
b=i+1;
}
while(sum!=0) /*当人数不为零时继续循环.*/
{
for(i=1;i<=num;i++) /*进行报数.*/
{
if(point>=sum) /*当前报数位置超过最后一人时从第一人报起.*/
{point=1;}
else point++;
}
num=a[point-1]; /*取密码.*/
code=b[point-1]; /*取号码.*/
for(i=point;i<=sum;i++) /*删去退出的人.*/
{
a[i-1]=a;
b[i-1]=b;
}
sum--; /*现存总人数.*/
flag++; /*退出的人数.*/
point--;
printf("已退出%d人,退出的人的编号为%d.\n",flag,code);
}
}
2.用循环链表
# include "stdio.h"
# include "alloc.h"
# define LEN sizeof(struct player)
struct player
{
int num; /*编号*/
int secret; /*密码*/
struct player *next;
};
int n; /*总人数*/
main()
{
int m;
void create();
void delete();
struct player *head;
head=(struct player *)malloc(LEN);
printf("请输入第一次的密码:\n");
scanf("%d",&m);
create(head);
delete(head,m);
}
void create(struct player *head)
{
struct player *p1,*p2,*p;
p1=p2=head;
printf("请输入编号,密码,当编号为零时输入结束.\n");
printf("总人数为:\n");
scanf("%d",&n);
do
{
p=p2;
p2=p1;
printf("编号:");
scanf("%d",&p2->num);
printf("密码:");
scanf("%d",&p2->secret);
p1=(struct player *)malloc(LEN);
p2->next=p1;
}while(p2->num!=0);
p->next=head;
free(p1);
}
void delete(struct player *head,int m)
{
int i;
struct player *p,*q;
int sum;
p=head;
sum=m;
while(n!=0)
{
for(i=1;i<sum;i++)
{ q=p;
p=p->next;
}
printf("%d,",p->num);
sum=p->secret;
q->next=p->next;
p=p->next;
n=n-1;
}
}
题目:有n个人围成一圈,顺序排号。从第一个人开始报数(从1到3报数),凡报到3的人退出圈子,问最后留下的是原来第几号的那位。
1. 程序分析:
2.程序源代码:
#define nmax 50
main()
{
int i,k,m,n,num[nmax],*p;
printf("please input the total of numbers:");
scanf("%d",&n);
p=num;
for(i=0;i<n;i++)
*(p+i)=i+1;
i=0;
k=0;
m=0;
while(m<n-1)
{
if(*(p+i)!=0) k++;
if(k==3)
{ *(p+i)=0;
k=0;
m++;
}
i++;
if(i==n) i=0;
}
while(*p==0) p++;
printf("%d is left\n",*p);
}
我写了一个带头结点的单链表逆置程序,程序的执行结果没什么问题,只是要高手指点一下我写的代码有没有哪里不规范的地方?
/****************************************************************
* 文件名:nz.c (nz就是逆置!)
* 文件描述:带头结点的单链表逆置
* 创建人:沁园枫
* 创建时间:2006年2月2日
****************************************************************/
#include "stdio.h"
typedef int ElemType
typedef struct node /*单链表结点*/
{
ElemType data; /*数据域*/
struct node *next; /*指针域*/
}slink;
void main() /*主函数*/
{
slink *creatslink(); /*声明xjlb(新建链表)函数*/
slink *inverse(slink *); /*声明nzlb(逆置链表)函数*/
slink *h,*p;
h= creatslink(); /*调用xjlb(新建链表)函数*/
p=h;
while(p-> next!=NULL)/*输出新建单链表后的各元素*/
{
printf("%d",p-> next ->data);
p=p-> next;
}
printf("\n");
h= inverse(h); /*调用nzlb(逆置链表)函数*/
while(h-> next!=NULL)/*输出逆置后的单链表各元素*/
{
printf("%d",h-> next ->data);
h=h-> next;
}
getch(); /*调用getch函数,不知道什么意思的自己查查!*/
}
slink * creatslink() /*新建链表函数*/
{
slink *h,*l,*a;
int i;
h=( slink *)malloc(sizeof(slink)); /*创建头结点*/
h-> next =NULL;
a=h;
for(i=0;i<5;i++)
{
l=( slink *)malloc(sizeof(slink));
l->data=i;
a-> next =l;
a=a-> next;
}
a-> next =NULL;
return h;
}
slink * inverse (slink *h) /*逆置链表函数*/
{
slink *p,*q;
p=h-> next;
h-> next =NULL;
while(p!=NULL)
{
q=p;
p=p-> next;
q-> next =h-> next;
h-> next =q;
}
return h;
}
单链表逆置算法
单链表逆置算法
struct node
{
int num;
struct node *next;
}
struct node* reverse(struct node *head)
//head 链表头结点 {
struct node *p,*temp1,*temp2;
if(head==NULL____①____) return head; //||head->next==NULL p=head->next;head->next=NULL;
while(____②____) //p!=NULL或p {
temp1=head;
____③____; //head=p; temp2=p;
p=p->next;
____④____; //temp2->next=temp1;或head->next=temp1; }//Match while statenment
return head; //返回逆置后的链表的头结点 }
struct node
{
int num;
struct node *next;
}
struct node* reverse(struct node *head)
//head 链表头结点 {
struct node *p,*temp1,*temp2;
if(head==NULL||head->next==NULL) return head;
p=head->next;head->next=NULL;
while(p!=NULL或p)
{
temp1=head;
head=p; temp2=p;
p=p->next;
temp2->next=temp1;或head->next=temp1; }//Match while statenment
return head; //返回逆置后的链表的头结点 }
最大子列和 n的阶乘中有多少个0 10进制to2进制单链表逆置判断链表中是否有环文件中有多少行
#ifndef MYLIST_H
#define MYLIST_H
#include <string>
struct listnode
{
int value;
listnode *next;
listnode ( int num ): value (num), next (NULL)
{
}
};
class Mylist
{
public:
Mylist ();
~Mylist ();
void Insert ( int value ) ;
std::string GetEntireList ();
void Reverse ();
private:
listnode *head;
};
#endif
//----------------------------------------------------------------
#include "Mylist.h"
using namespace std;
Mylist::Mylist ()
{
head = NULL;
}
Mylist::~Mylist()
{
if ( head != NULL )
{
listnode *temp = head ;
while ( head != NULL )
{
temp = head->next;
delete head;
head = temp;
}
}
}
void Mylist::Insert ( int value )
{
if ( head == NULL )
{
head = new listnode ( 0 );
}
listnode *curnode = new listnode ( value );
listnode *temp = head;
while ( temp->next != NULL )
{
temp = temp->next;
}
temp->next = curnode;
}
string Mylist::GetEntireList ()
{
string str = "";
listnode *temp = head->next ;
while ( temp != NULL )
{
str += temp->value + '0' ;
str += ' ' ;
temp = temp->next;
}
return str;
}
void Mylist::Reverse ()
{
if ( head == NULL || head->next == NULL )
{
return;
}
listnode *end = head;
while ( end->next != NULL )
{
end = end->next;
}
listnode *curnode = head->next;
while ( curnode != end )
{
// 将curnode从链表中删除
head->next = curnode->next;
//将curnode插入到end后面
curnode->next = end->next;
end->next = curnode;
curnode = head->next;
}
}
// 判断链表中是否存在着环
bool Mylist::HasaCycle ()
{
listnode *itr1 = NULL;
listnode *itr2 = NULL;
// 如果链表为空则返回false
if ( head == NULL || head != NULL && ( head->next == NULL ) )
{
return false;
}
itr1 = head->next;
itr2 = itr1->next;
if ( itr2 == NULL )
{
return false;
}
// 如果head-〉next指向了head ,或者itr1-〉head 指向了itr1,或者itr1-〉next指向了head,有环
if ( itr1 == head || itr2 == itr1 || itr2 == head )
{
return true;
}
// head -> itr1 -> itr2 -> *
while ( itr1 != NULL && itr2 != NULL && itr2->next != NULL )
{
if ( itr1 == itr2 || itr2->next == itr1 ) 记号
{
return true;
}
itr1 = itr1->next;
itr2 = itr2->next->next;
}
return false;
}
}
//================================
// test。cpp
#include <fstream>
#include <iostream>
#include <math.h>
#include <string>
#include <list>
#include "Mylist.h"
using namespace std;
//==================================================
// 获得data。txt文件中有多少行
int GetTotalLineCount ()
{
ifstream ifs;
ifs.open ( "data.txt", ios::in );
int count = 1 ;
//string p;
// while ( getline( ifs, p ) ) { ++count; }
char ch;
while ( ifs.get( ch ) )
{
if ( ch == '\n')
{
++count;
}
}
ifs.close();
return count;
}
//================================================================
//计算正整数num的阶乘中有多少个0
const int FIVE = 5;
// 得到不大于 num的5的最大次方,如 num = 5 时返回 1, 26 时返回 2
int LowerNearestPow ( int num )
{
int powoffive = 1;
while ( pow ( FIVE, powoffive ) <= num )
{
++powoffive;
}
return powoffive;
}
// 考虑25,125, 625 等的情况
int GetZerosofXFactorial ( int num )
{
int lowernearestpow = LowerNearestPow ( num );
int zerocount = 0;
while ( lowernearestpow > 0 )
{
zerocount += num / pow ( FIVE, lowernearestpow ) ;
--lowernearestpow;
}
return zerocount;
}
//=========================================================================
// 将10进制正整数num转化为2进制数
const int TWO = 2 ;
string DecimalToBin ( unsigned int num )
{
if ( num == 0 )
{
return "0";
}
string binform = "";
while ( num > 0 )
{
binform.insert ( binform.begin(), num % TWO + '0' );
num /= TWO ;
}
return binform;
}
//=========================================================================
// 求一个整数数组的最大子列和
int MaxSubArraySum ( int *array, int size )
{
int sum = 0;
while ( size >= 0 )
{
if ( array[size] > 0 )
{
sum += array[ size ];
}
else if ( array[size] < 0 )
{
if ( sum == 0 )
{
sum = array[size];
}
}// end of else
--size;
}// end of while
return sum;
}
//=========================================================================
void main ()
{
// int count = GetTotalLineCount ();
// cout << count << endl;
// while ( 1 )
// {
// cout << " 请输入一个数,本程序将计算这个数的阶乘中有多少个0:" << endl;
// int num;
// cin >> num ;
// cout << GetZerosofXFactorial ( num ) << endl;
// }
// while ( 1 )
// {
// cout << " 请输入一个正整数,本程序将计算这个数的二进制形式:" << endl;
// unsigned int num;
// cin >> num ;
// cout << DecimalToBin ( num ) << endl;
// }
// Mylist intlist;
// for ( int i=0; i<8; ++i )
// {
// intlist.Insert ( i );
// }
// string str = intlist.GetEntireList ();
// cout << " 在逆置之前链表内容为:" << str << endl;
// intlist.Reverse();
// str = intlist.GetEntireList ();
// cout << " 在逆置之后链表内容为:" << str << endl;
// int sum[] = { -1, -2, 0, -4, -6, -1}; // -1
//{ 1, 2, -1, -4, 6, 1}; // 10
// int size = sizeof( sum ) / sizeof(int) - 1;
// cout << MaxSubArraySum ( sum , size ) << endl;
}
又n个人围成一圈,顺序排号.从第一个人开始报数(从1到m报数),凡报到m的人退出圈子,问最后留下的是原来第几号?链表:
#include<stdio.h>
#include<malloc.h>
typedef struct node{
int info;
struct node *next;
};
node *tbuildhlink(int n) /*带头节点的尾插法*/
{
node *head,*s,*p2;
int i=1;
head=(node *)malloc(sizeof(node));
p2=head;
while(i<=n)
{
s=(node *)malloc(sizeof(node));
s->info=i;
p2->next=s;
p2=s;
i++;
}
if(p2) p2->next=head;
return(head);
}
void Display(struct node* head)
{
node *p;
p=head->next;
if(!p)
{
printf("\nthe hlink is empty!");
}
else
{
printf("\nthe value of the hlink is:\n");
while(p!=head)
{
printf("%d--->",p->info);
p=p->next;
}
}
printf("^\n");
}
int delete_node(struct node *head,int n,int m)
{
int count=1,sum=n;
struct node *p,*pre;
pre=head;
p=pre->next;
while(sum>1)
{
if(p==head)
{
p=p->next;
}
if(count<m)
{
pre=p;
p=p->next;
count++;
}
if(count==m)
{
if(p==head)
{
p=p->next;
}
printf("第%d个人出列.\n",p->info);
pre->next=p->next;
free(p);
p=pre->next;
count=1;
sum--;
}
}
return(pre->info);
}
int main()
{
node *head;
int n,m;
printf("输入n,m:");
scanf("%d%d",&n,&m);
head=tbuildhlink(n);
Display(head);
printf("最后剩下第%d个.\n",delete_node(head,n,m));
return(0);