数据结构使用教程 代码

数据结构和算法实践

  1 //2.4-2,3,5
  2 //get max value in linked list
  3 ElemType MaxValue(LNode * head)
  4 {
  5   if(head==NULL)
  6   {
  7       cerr<<"Linked list is empty!"<<endl;
  8       exit(1);
  9   }
 10   ElemType max=head->data;
 11   LNode* p=head->next;//用temp指针p指向链表
 12   while(p!=NULL)
 13   {
 14      if(p->data>max)
 15          max=p->data;
 16      p=p->next; //指向下一个指针
 17   }
 18   return max;
 19 }
 20 
 21 //
 22 int Count(LNode* head,ElemType x)
 23 {
 24   int n=0;
 25   while(head!=NULL)
 26   {
 27       if(head->data==x)
 28           n++;
 29       head=head->next;
 30   }
 31   return n;
 32 }
 33 
 34 LNode* Merge(LNode* &p1,LNode* &p2)
 35 {
 36     LNode a; //便于处理，创建结果有序表头附加节点，
 37     LNode* p=&a;//p指向结果链表的表尾节点，初始指向表头附加节点
 38     p->next=NULL;
 39     while(p1!=NULL&&p2!=NULL)
 40     {
 41         if(p1->data<=p2->data)
 42         {
 43             p->next=p1;
 44             p1=p1->next;
 45         }else{
 46             p->next=p2;
 47             p2=p2->next;
 48         }
 49         p=p->next;
 50     }
 51 
 52     if(p1!=NULL) p->next=p1;
 53     if(p2!=NULL) p->next=p2;
 54 
 55     p1=p2=NULL;
 56     return a.next;
 57 }
 58 
 59 template<class ElemType>
 60 class Stack{
 61     ElemType *stack;
 62     int top;
 63     int MaxSize;
 64 public:
 65     Stack();
 66     Stack(Stack& s);
 67     Stack& operator=(Stack& s);
 68     void Push(ElemType item);
 69     ElemType Pop();
 70     ElemType Peek();
 71     bool EmptyStack();
 72     ~Stack();
 73 };
 74 
 75 template<class ElemType>
 76 ElemType Stack<ElemType>::Pop()
 77 {
 78     ///write something
 79 }
 80 
 81 template<class ElemType>
 82 ElemType Stack<ElemType>::Peek()
 83 {
 84     ///write something
 85 }
 86 
 87 template<class ElemType>
 88 void Stack<ElemType>::Push(ElemType item)
 89 {
 90     ///write something
 91 }
 92 
 93 template<class ElemType>
 94 bool Stack<ElemType>::EmptyStack()
 95 {
 96     return top==-1;
 97 }
 98 //4.3 -2,3
 99 //后缀表达式的求值定义 ，操作数栈为引用参数
100 double Compute(Stack<double> &S,char* str)
101 {
102   double x,y;// 保存浮点数
103   int i=0;
104   while(str[i]){
105       if(str[i]=='')
106       {
107           i++;continue;
108       }
109       switch(str[i]){
110        case '+':
111            x=S.Pop()+S.Pop();
112            i++;
113            break;
114        case '-':
115            x=S.Pop();
116            x=S.Pop()-x;
117            i++;
118            break;
119        case '*':
120            x=S.Pop()*S.Pop();
121            i++;
122            break;
123        case '/':
124            x=S.Pop();
125          if(x!=0.0) x=S.Pop()/x;
126          else
127          {
128             cerr<<"Divide by zero!"<<endl;
129             exit(1);
130          }
131          i++;
132          break;
133        default:
134            x=0;
135            while(str[i]>=48&&str[i]<=57){
136               x=x*10+str[i]-48;i++;
137            }
138            if(str[i]=='.'){
139                i++;
140                y=0;
141                double j=10.0;
142                while(str[i]>=48&&str[i]<=57){
143                    y=y+(str[i]-48)/j;
144                    i++;j*=10;
145                }
146                x+=y;
147            }
148       }
149       S.Push(x);
150   }
151   if(S.EmptyStack())
152   {
153       cerr<<"Stack is empty!"<<endl;
154       exit(1);
155   }
156   x=S.Pop();
157   if(S.EmptyStack()) return x;
158   else
159   {
160       cerr<<"expression error!"<<endl;
161       exit(1);
162   }
163 }
164 
165 //返回运算符op的优先级数值
166 int Precedence(char op)
167 {
168    switch(op)
169    {
170      case '+':
171      case '-':
172          return 1;
173      case '*':
174      case '/':
175          return 2;
176      case '(':
177      case '@':
178      default:
179          return 0;
180    }
181 }
182 
183 //中缀表达式转换成后缀表达式 运算符栈为引用参数
184 void Change(Stack<char>& R,char* s1,char* s2)
185 {
186    R.Push('@');
187    int i=0,j=0;
188    char ch=s1[i];
189    while(ch!='\0')
190    {
191        if(ch=='') ch=s1[++i];
192        else if(ch=='('){
193            R.Push(ch);ch=s1[++i];
194        }
195        else if(ch==')'){
196            while(R.Peek()!='(') {s2[j++]=R.Pop();}
197            R.Pop(); // 删除站顶的左括号
198            ch=s1[++i];
199        }
200        else if(ch=='+'||ch=='-'||ch=='*'||ch=='/'){
201 
202            char w=R.Peek();
203            while(Precedence(w)>=Precedence(ch))
204            {
205                s2[j++]=w;R.Pop();
206                w=R.Peek();
207            }
208            R.Push(ch);
209            ch=s1[++i];
210        } else{
211            if((ch<'0'||ch>'9')&&ch!='.'){
212                cout<<"error"<<endl;
213                exit(1);
214            }
215            while((ch<'0'||ch>'9')&&ch!='.'){
216                s2[j++]=ch;
217                ch=s1[++i];
218            }
219            s2[j++]='';
220        }
221 
222        ch=R.Pop();
223        while(ch!='@'){
224            if(ch=='(')
225            {
226                cerr<<"expression error"<<endl;exit(1);
227            }else{
228                s2[j++]=ch;
229                ch=R.Pop();
230            }
231        }
232        s2[j++]='\0';
233    }
234 }
235 
236 //利用辗转相除和递归的方法求两个整数的最大公约数
237 int f1(int a,int b)
238 {
239   //cout<<"a="<<a<<","<<"b="<<b<<endl;
240   if(a%b==0) return b;
241   else return f1(b,a%b);
242 }
243 
244 //Fibonacci 数列递归和非递归算法
245 int Fib(int n)
246 {
247     if(n==1||n==2) return n-1; //终止递归条件
248     else return Fib(n-1)+Fib(n-2);
249 }
250 //时间复杂度O(2*n) 空间复杂度O(n)
251 
252 int Fib2(int n)
253 {
254     int a,b,c;
255     a=0;b=1;
256     if(n==1||n==2) return n-1;
257     else
258     {
259         for(int i=3;i<=n;i++)
260         {
261             c=a+b;
262             a=b;
263             b=c;
264         }
265         return c;
266     }
267 }
268 //时间复杂度O(n) 空间复杂度O(1)
269 
270 //编写汉诺塔问题的非递归算法
271 void Hanoi1(int n,int a,int b,int c)
272 {
273   struct items{
274       int n,a,b,c;
275   };
276   items s[10]; //used for sequence stack
277   int top=-1;
278   items temp;
279   while(top!=-1||n>=1)
280   {
281      while(1){
282          if(n==1) {cout<<a<<"=>"<<c<<endl;n=0;break;}
283          temp.n=n;temp.a=a;temp.b=b;temp.c=c;
284          top++;s[top]=temp; //enter stack
285          cout<<"n1="<<n<<endl;
286          n=n-1;int d=b;b=c;c=d;//a's value no change
287          cout<<"n2="<<n<<endl;
288      }
289      if(top!=-1){
290        temp=s[top];top--;
291        n=temp.n;a=temp.a;b=temp.b;c=temp.c;
292        cout<<a<<"->"<<c<<endl;
293        n=n-1;int d=a;a=b;b=d;//c's value no change
294      }
295   }
296 }
297 
298 struct BTreeNode{
299     ElemType data;
300     BTreeNode* left;
301     BTreeNode* right;
302 };
303 
304 // binary tree node count
305 int BTreeNodeCount(BTreeNode* BT)
306 {
307   if(BT==NULL) return 0;
308   else
309       return BTreeNodeCount(BT->left)+BTreeNodeCount(BT->right)+1;
310 }
311 
312 // binary tree leaf count
313 int BTreeLeafCount(BTreeNode* BT)
314 {
315    if(BT==NULL) return 0;
316    else if(BT->left==NULL&&BT->right==NULL) return 1;
317    else return BTreeLeafCount(BT->left)+BTreeLeafCount(BT->right);
318 }