[Algorithm]栈和队列

一.栈和队列综合(算法)

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1.判断单链表(带头结点)的结点值(字符型)是否中心对称

 1 bool IsSymmetry( LinkList& L )
 2 {
 3     char S[MAXSIZE];
 4     int top = -1, len = 0, i;
 5     LinkList p;
 6     p = L->next;
 7     while ( p ) { p = p->next; len++; }
 8     p = L->next;
 9     for (i=0;i<len/2;i++)
10     {
11         S[++top] = p->data;
12         p = p->next;
13     }
14     i--;
15     if ( len % 2 )
16         p = p->next;
17     while ( top != -1 )
18     {
19         if ( p->data != S[top] )
20             return false;
21         top--;
22         p = p->next;
23     }
24     return true;
25 }

2.共享栈由两个顺序栈S1,S2构成,总大小为100,请设计S1,S2入栈,出栈的算法

 1 #define MAXSIZE 100
 2 ElemType S[MAXSIZE];
 3 int top[2] = { -1,MAXSIZE };
 4 bool Push( int i, ElemType x )
 5 {
 6     if ( i < 0 || i>1 || top[1] - top[0] == 1 )
 7         return false;
 8     if ( i == 0 ) S[++top[0]] = x;
 9     else S[--top[1]] = x;
10     return true;
11 }
12
13 bool Pop( int i, ElemType x )
14 {
15     if ( i < 0 || i>1
16          || ( i == 0 && top[0] == -1 )
17          || ( i == 1 && top[1] == MAXSIZE ) )
18         return false;
19     if ( i == 0 ) x = S[top[0]--];
20     else x = S[top[1]++];
21     return true;
22 }

3.如果希望循环队列中的元素都能得到利用,则需设置一个标志域tag,并以tag的值为0或1来区分队头指针front和队尾rear相同时的队列状态是"空"还是"满",编写与此结构相应的入队和出队算法

 1 ElemType Q[MAXSIZE];
 2 int front = -1, rear = -1;
 3 // 队空条件:    front==rear&&tag==0
 4 // 队满条件:    front==rear&&tag==1
 5 // 进队操作:    rear=(rear+1)%MAXSIZE;
 6 //            Q[rear]=x;
 7 //            tag=1;
 8 // 出队操作:    front=(front+1)%MAXSIZE;
 9 //            x=Q[front];
10 //            tag=0;
11
12 // 1)"tag"法循环队列入队算法
13 bool EnQueue( ElemType x )
14 {
15     if ( front == rear && tag == 1 )
16         return false;
17     rear = ( rear + 1 ) % MAXSIZE;
18     Q[rear] = x;
19     tag = 1;
20     return true;
21 }
22
23 // 2)"tag"法循环队列出队算法
24 bool DeQueue( ElemType& x )
25 {
26     if ( front == rear && tag == 0 )
27         return false;
28     front = ( front + 1 ) % MAXSIZE;
29     x = Q[front];
30     tag = 0;
31     return true;
32 }

4.Q是一个队列,S是一个空栈,实现将队列中的元素逆置的算法

 1 ElemType S[MAXSIZE], Q[MAXSIZE];
 2 int top = -1, front = -1, rear = -1;
 3 void Inverse(ElemType S[], ElemType Q[])
 4 {
 5     ElemType x;
 6     while ( front != rear )
 7     {
 8         x = Q[++front];
 9         S[++top] = x;
10     }
11     while ( top != -1 )
12     {
13         x = S[top--];
14         Q[++rear] = x;
15     }
16 }

5.利用两个栈S1,S2模拟一个队列,已知栈的4个运算如下:

 1 // 已知:
 2 void Push(Stack& S, ElemType x);
 3 void Pop(Stack& S, ElemType& x)
 4 bool IsEmpty(Stack& S);
 5 bool IsOverflow( Stack& S );
 6
 7 bool EnQueue( Stack& S1, Stack& S2, ElemType x )
 8 {
 9     if ( !IsOverflow( S1 ) )
10     {
11         Push( S1, x );
12         return true;
13     }
14     if ( !IsEmpty( S2 ) )
15         return false;
16     while (!IsEmpty(S1))
17     {
18         Pop( S1, t );
19         Push( S2, t );
20     }
21     Push( S1, x );
22     return true;
23 }
24
25 bool DeQueue( Stack& S1, Stack& S2, ElemType& x )
26 {
27     if (!IsEmpty(S2))
28     {
29         Pop( S2, x ); return true;
30     }
31     if ( IsEmpty( S1 ) )
32         return false;
33     while (!IsEmpty(S1))
34     {
35         Pop( S1, t );
36         Push( S2, t );
37     }
38     Pop( S2, x );
39     return true;
40 }
41
42 bool IsEmpty( Stack& S1, Stack& S2 )
43 {
44     if ( IsEmpty( S1 ) && IsEmpty( S2 ) )
45         return true;
46     return false;
47 }

6.括号匹配问题:判别表达式中括号是否匹配(只含有$(),[],\{\}$)

 1 bool IsBracketMatch( char*str )
 2 {
 3     char S[MAXSIZE];
 4     int top = -1;
 5     for ( int i = 0; str[i]; i++ )
 6     {
 7         char c = str[i];
 8         switch ( c )
 9         {
10         case '(':
11         case '[':
12         case '{':
13             S[++top] = c;
14             break;
15         case ')':
16             c = S[top--];
17             if ( c != '(' )return false;
18             break;
19         case ']':
20             c = S[top--];
21             if ( c != '[' )return false;
22             break;
23         case '}':
24             c = S[top--];
25             if ( c != '{' )return false;
26             break;
27         default:
28             break;
29         }
30     }
31     return top == -1;
32 }

7.利用栈实现以下递归函数的非递归计算:

$$P_{n}(x)=
\cases{
1 & n=0\cr
2x & n=1\cr
2x\cdot{P_{n-1}(x)}-2(n-1)\cdot{P_{n-2}(x)} & n>1
}$$

 1 double P( int n, double x )
 2 {
 3     struct Stack
 4     {
 5         int n;        // 层
 6         double val;    // 数值结果
 7     }S[MAXSIZE];
 8     int top = -1, fv1 = 1, fv2 = 2 * x;
 9     for ( int i = n; i > 1; i-- )
10         S[++top].n = i;
11     while ( top != -1 )
12     {
13         S[top].val = 2 * x*fv2 - 2 * ( S[top].n - 1 )*fv1;
14         fv1 = fv2;
15         fv2 = S[top--].val;
16     }
17     if ( n == 0 ) return fv1;
18     return fv2;
19 }