数据结构复习笔记(6)

1，  最大子序列和问题(四种解法)

 
  class  Test
     {
        public static void Main()
        {
            int[] a = {-2,11,-4,13,-5,-2};
            int result = MaxSubsequenceSumFour(a,6);
            Console.WriteLine(result);
            Console.Read();

        }
        public static int MaxSubsequenceSumOne(int[] A,int n)
        { //算法1
            int CurSum,MaxSum=0,i,j;
            for(i=0;i<n;i++)
            {
                CurSum = 0;
                for(j=i;j<n;j++)
                {
                    CurSum += A[j];
                    if(CurSum > MaxSum)
                    {
                        MaxSum = CurSum;    
                    }
                }
                
            }    
            return MaxSum;
        }
        public static int MaxSubsequenceSumTwo(int[] A,int n)
        {//算法2
            int CurSum,MaxSum=0,i,j,k;
            for(i=0;i<n;i++)
                for(j=i;j<n;j++)
                {
                    CurSum = 0;
                    for(k=i;k<=j;k++)
                        CurSum += A[k];
                    if(CurSum > MaxSum)
                        MaxSum = CurSum ;
                }
            return MaxSum;
        }

        public static int MaxSubsequenceSumThree(int[] A,int n)
        {//算法3
             return MaxSubSum( A, 0, n - 1 );
        }

        public static int MaxSubsequenceSumFour(int[] A,int n)
        {//算法4
            int ThisSum, MaxSum, j;
            ThisSum = MaxSum = 0;
            for( j = 0; j < n; j++ )
            {
                ThisSum += A[ j ];
                if( ThisSum > MaxSum )
                    MaxSum = ThisSum;
                else if( ThisSum < 0 )
                    ThisSum = 0;
            }
            return MaxSum;
        }


        public static int Max3( int A, int B, int C )
        {
            return A > B ? A > C ? A : C : B > C ? B : C;
        }

        public static int MaxSubSum(int[ ] A, int Left, int Right )
        {
            int MaxLeftSum, MaxRightSum;
            int MaxLeftBorderSum, MaxRightBorderSum;
            int LeftBorderSum, RightBorderSum;
            int Center, i;
            if( Left == Right )  /**//* Base case */
                if( A[ Left ] > 0 )
                    return A[ Left ];
                else
                    return 0;
            Center = ( Left + Right ) / 2;
            MaxLeftSum = MaxSubSum( A, Left, Center );
            MaxRightSum = MaxSubSum( A, Center + 1, Right );
            MaxLeftBorderSum = 0;
 LeftBorderSum = 0;
            for( i = Center; i >= Left; i-- )
            {
                LeftBorderSum += A[ i ];
                if( LeftBorderSum > MaxLeftBorderSum )
                    MaxLeftBorderSum = LeftBorderSum;
            }
            MaxRightBorderSum = 0;
 RightBorderSum = 0;
            for( i = Center + 1; i <= Right; i++ )
            {
                RightBorderSum += A[ i ];
                if( RightBorderSum > MaxRightBorderSum )
                MaxRightBorderSum = RightBorderSum;
            }
            return Max3( MaxLeftSum, MaxRightSum,MaxLeftBorderSum + MaxRightBorderSum );
        }
    
    }

2，  高效率的取幂运算

class  Test
     {
        public static void Main()
        {
            Console.WriteLine(Pow( 3, 4 ));

            Console.Read();

        }
        static long Pow( int X, int N )
        {
            if( N == 0 )
                return 1;
            if( N == 1 )
                return X;

            if(N%2==0) //偶数
                 return Pow( X * X, N / 2 );
            else
                 return Pow( X * X, N / 2 ) * X;
        }
        static long PowTwo( int X, int N )
        {
            if( N == 0 )
                return 1;
            if(N%2==0)
                 return Pow( X * X, N / 2 );
            else
                 return Pow( X , N -1 ) * X;
        }
}

3，一元多项式运算（链表表示）

＃include < stdio.h >
＃include < malloc.h >
＃include < stdlib.h >
#define   LEN  sizeof(node) 

typedef  struct  polynode
{
 int coef;//系数
 int exp;//指数
 struct polynode *next;
} node;

node  *  create( void )
{
 node *h,*r,*s;
 int c,e;
 h=(node *)malloc(LEN);//头结点
 r=h;
 printf("coef:");
 scanf("%d",&c);
 printf("exp: ");
 scanf("%d",&e);
 while(c!=0)
 {
  s=(node *)malloc(LEN);
  s->coef=c;
  s->exp=e;
     r->next=s;
     r=s;
     printf("coef:");
     scanf("%d",&c);
     printf("exp: ");
     scanf("%d",&e);
 }
 r->next=NULL;//表尾
 return(h);
}
 
polyadd(node  * polya, node  * polyb)
{
 node *p,*q,*pre,*temp;
 int sum;
 p=polya->next;
 q=polyb->next;
 pre=polya;
 while(p!=NULL&&q!=NULL)
 {
  if(p->exp<q->exp)
  {
   pre->next=p;pre=pre->next;
   p=p->next;
  }
  else if(p->exp==q->exp)
  {
   sum=p->coef+q->coef;
   if(sum!=0)
   {
    p->coef=sum;
    pre->next=p;pre=pre->next;
    p=p->next;temp=q;q=q->next;free(temp);
   }
   else
   {temp=p->next;free(p);p=temp;
    temp=q->next;free(q);q=temp;
   }
     }
 else 
 {pre->next=q;pre=pre->next;
  q=q->next;
 }
    }
if(p!=NULL)
pre->next=p;
else
pre->next=q;                                                         
}

void  print(node  *  p)
{                                                                                                      
    while(p->next!=NULL)
    {   
        p=p->next;
        printf("     %d*x^%d",p->coef,p->exp);   
           
    }
}    
main()
{
 node * polya,* polyb;
 printf("Welcome to use!\n");
 printf("\nPlease input the ploya include coef && exp:\n");
 polya=create();
 print(polya);
 printf("\nPlease input the ployb include coef && exp:\n");
 polyb=create();
 print(polyb);
    printf("\nSum of the poly is:\n");
    polyadd(polya,polyb);
    print(polya);
    printf("\n");
}  
datastruct.h

typedef  struct  list
{
    int c;    //多项式的项数
    int e;    //多项式的指数 
    struct list *next;    //下一结点
} ;
typedef  struct  list  * LinkList;
typedef  struct  list Node;

另一个示例:

// File: ListOper.h

#ifndef DATASTRUCT
#define  DATASTRUCT
#include  " datastruct.h "
#endif
// some functions declaretioin
bool  CreateList(LinkList  & L);
Node  * CreateNode( int  e,  int  c);
void  FreeList(LinkList  & L);
void  SortList(LinkList  & L);
void  DeleteNextNode(Node  * d);
void  SweepNextNode(Node  * s);
void  OutPutList(LinkList  & L);

相关函数的实现，对应文件ListOper.cpp:
// File: ListOper.cpp
#include  < stdlib.h >
#include  < iostream >
#ifndef DATASTRUCT
#define  DATASTRUCT
#include  " datastruct.h "
#endif
#include  " ListOper.h "
using   namespace  std;
bool  CreateList(LinkList  & L)
{
    //TODO: 创建线性链表
    Node *head;
    head=(Node *)malloc(sizeof(Node));
    if(head==NULL)
    {
        cout << "内存分配错误" << endl;
        return false;
    }
    head->next=NULL;
    head->c=0;
    head->e=0;
    L=head;
    return true;
}
Node  * CreateNode( int  e,  int  c)
{
    //TODO: 创建结点
    Node * pos;
    pos=(Node *)malloc(sizeof(Node));
    if(pos==NULL)
    {
        cout << "内存分配错误" << endl;
        exit(1);
    }
    pos->e=e;
    pos->c=c;
    pos->next=NULL;
    return pos;
}
         
void  FreeList(LinkList  & L)
{
    //TODO: 释放整个线性表所占用的内存空间
    Node *pos;
    Node *next;
    pos=L;
    while(pos!=NULL)
    {
        next=pos->next;
        free(pos);
        pos=next;
    }
}
void  SortList(LinkList  & L)
{
    bool flag=true; //是否需要排序标志
    Node *head=L->next;
    Node *pos;
    Node *last;
    Node *temp;
    if(head->next==NULL)
    {
        return;
    }
    while(flag)
    {
        flag=true;
        last=head;
        pos=last->next;
        if(last==NULL||last->next==NULL)
        {
            break;
        }
        while(last!=NULL && last->next!=NULL)
        {
            flag=false;
            pos=last->next;
            if(last->e<pos->e)    //哈哈哈哈哈，HTML代码
            {
                SweepNextNode(last);
                flag=true;
            }
            if(last->e==pos->e)
            {
                last->c+=pos->c;
                DeleteNextNode(last);
                flag=true;
                /**//*last=last->next;
                pos=last->next;*/
            }
            last=last->next;
        }
    }
}
void  DeleteNextNode(Node  * d)
{
    Node *temp;
    temp=d->next;
    d->next=temp->next;
    free(temp);
}
void  SweepNextNode(Node  * s)
// 一点偷懒的办法，只交换值，不修改指针
{
    int c,e;
    c=s->c;e=s->e;
    s->c=s->next->c;s->e=s->next->e;
    s->next->c=c;s->next->e=e;
}
void  OutPutList(LinkList  & L)
{
    Node *pos;
    pos=L->next;
    cout << "输出表达式：";
    while(pos!=NULL)
    {
        if(pos->c>0)
        {
            cout << "+";
        }
        if(pos->c!=1)
        {
            cout << pos->c;
        }
        if(pos->e!=0)
        {
            cout << "x^";
            cout << pos->e;
        }
        pos=pos->next;
    }
    cout << endl;
}

主单元文件main.cpp:
#include  < iostream >
#include  < stdlib.h >
#include  < ctype.h >
#include  " ListOper.h "
using   namespace  std;
LinkList AnayString( char  aString[],  int  aLength);
int  main( int  argc,  char   * argv[])     // -------------------------------
{
    LinkList L;
    char InStr[1024];
    int len;
    cout << "一元稀疏多项式计算器" << endl;
  cout << "请输入一个1024个字符以内的稀疏多项式：";
    cin >> InStr;
    len=strlen(InStr);
    L=AnayString(InStr,len);
    SortList(L);
    OutPutList(L);
    FreeList(L);
    system("PAUSE"); 
    return 0;
}
LinkList AnayString( char  aString[],  int  aLength)     // ---------------
// TODO: 字符串分析函数 
{
    LinkList L=NULL;
    Node *pos=NULL;
    Node *last;
    Node *head;
    CreateList(L);
    head=L;
    last=head;
    int c=0;
    int e=0;
    char temp[1];
    char tp;
    bool plus=true;
    char status='n';    //状态指示符,我省略了系数为负的情况
    /**//*
    n: 非运算状态
    c: 正在计算系数
    e: 正在计算指数 
    p: 指数为0 
    f: 完成了一个项目的输入
    */
    for(int i=0;i<aLength;i++)
    {
        temp[0]=aString[i];
        tp=temp[0];
        switch(status)
        {
            case 'n':
            {
                c=0;e=0;
                status='c';
                if(tp=='-')
                {
                    plus=false;
                    continue;
                }
                if(isdigit(tp))
                {
                    c=atoi(temp);
                    continue;
                }
                if(tp=='x')//多项式以x开头
                {
                    c=1;
                    status='e';
                    continue;
                }
            }
            case 'c':
            {
                if(isdigit(aString[i]))
                {
                    if(plus)
                    {
                        c=c*10+atoi(temp);
                    }
                    else
                    {
                        c=c*10-atoi(temp);
                    }
                    continue;
                }
                if(tp=='x')
                {
                    if(c==0)
                    {
                        c=1;
                    }
                    status='e';
                    e=0;
                    continue;
                }
                //此处考虑了常数项出现在其他位置的可能
                if(tp=='+')
                {
                    plus=true;
                    status='p';
                    continue;
                }
                if(tp=='-')
                {
                    plus=false;
                    status='p';
                    continue;
                }
                /**//*if(temp[0]=='^')
                {
                    status='e';
                    e=0;
                    continue;
                }*/ //此种情况不可能出现
                continue;
            }    //正在解析系数
            case 'e':
            {
                if(tp=='^')
                {
                    continue;
                }
                if(isdigit(tp))
                {
                    e=e*10+atoi(temp);
                    continue;
                }
                if(tp=='+')
                {
                    plus=true;
                    status='f';
                    continue;
                }
                if(tp=='-')
                {
                    plus=false;
                    status='f';
                    continue;
                }
            }            //正在解析系数
            case 'p':
            {
                e=0;
                status='f';
                continue;
            }
            case 'f':
            {
                pos=CreateNode(e,c);
                last->next=pos;
                last=pos;
                c=0;e=0;
                status='c';
                i--;
                continue;
            }
        }
    }    
    pos=CreateNode(e,c);
    last->next=pos;
    return L;
}

4，  中缀表达式转换为后缀表达式

#include < iostream.h >

const   int  MAX = 40 ;

void  main( void ) {
    char infix[MAX]={'#'};
    char oprator[MAX]={'@','#'};
    int opr=1;
    char postfix[12]={'#'};
    int post=0;
    int i,j,cnt=0,cntl;
    char c;

    //输入表达式，以等号结束
    cin.get(c);
    while(c!='='){
        infix[cnt]=c;
        cnt++;
        cin.get(c);
    }
    cntl=cnt;

    for(i=0;i<cnt;i++){
        switch(infix[i]){
        //左括号就直接入栈
        case '(':
            cntl=cntl-2;
            oprator[opr]=infix[i];
            opr++;
            break;
        //右括号则先退栈，直到遇见第一个左括号
        case ')':
            for(j=opr-1;j>0;j--){
                if(oprator[j]!='('){
                    postfix[post]=oprator[j];
                    oprator[j]='#';
                    post++;
                }
                else{
                    oprator[j] = '#';
                    break;
                }
            }
            opr=j;
            break;
        case '*':
        case '/':
            //如果前一个运算符为*或/，则先退栈,再入栈,否则直接入栈
            if (oprator[opr] == '*' || oprator[opr] == '/') {
                postfix[post] = oprator[opr];
                oprator[opr]='#';
                post++;
            }
            oprator[opr] = infix[i];
            opr++;
            break;
        case '+' :
        case '-' :
            //如果上一个运算符不是左括号也不是栈顶，则先退栈再入栈
            if (oprator[opr-1] != '(' && oprator[opr-1] != '@') {            
                postfix[post] = oprator[opr];
                oprator[opr]='#';
            }
            oprator[opr] = infix[i];
            opr++;
            break;
        default :
            //如果是数字则直接进入后缀表达式数组
            postfix[post] = infix[i];
            post++;
            break;
        }
    }

    //如果扫描完成，则退栈
    for(j=opr-1;j>0;j--){
        if(oprator[j]!='@'){
            postfix[post]=oprator[j];
            oprator[j]='#';
        }
        else
            break;
    }

    //输出结果
    for(i=0;i<cntl;i++)
        cout << postfix[i];
    cout << endl;
}