刚才发表了一篇后缀表达式求值(逆波兰表达式求值)的代码,虽然其在visual C++2010中编译出现了问题,不过幸好有好心人的热心帮助解决了。
那么为什么要用后缀表达式求值?根据我自己的理解,后缀表达式求值不必判断运算符的优先级,只要按顺序执行。每遇到一个操作符就处理一次,便于计算机处理。
那么下面我就分享一下我的中缀表达式转后缀的代码:
也许大家在考试时转化一定会:
中缀表达式:(8+9*10)-4/2+3
其转化思路:
1、将每个操作符对应的两个操作数用括号括上(((8+(9*10))-(4/2))+3)
2、将操作符移到对应的括号外(((8(910*)+)(42)/)-3)+
3、去掉括号即可 8910*+42/-3+
如果要用数据结构中的栈和队列实现
1、用一个字符串存储表达式
2、扫描字符串。当其为0--9时直接入队列,遇到左括号入栈,操作符级别 #:0用于最后判断,+ -:1,* /:2
3、首先向堆中放一个#,当进入堆的操作符的级别不小于栈顶操作符的优先级,则入栈,否则弹出栈顶元素并进入队列,将下一个操作符压入栈。
4、一直循环直到将所有字符处理完
5、最后将所有元素出队列并打印就可得到后缀表达式
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#include<stdio.h>
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#define Max 20
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//自定义栈
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template<class Elem>
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struct Stack
{
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int top;
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Elem *p;
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int size;
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};
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template<class Elem>
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void Set_Ssize(Stack<Elem> &sta,int n)
{
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sta.size=n;
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sta.p=new Elem[sta.size];
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sta.top=0;
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}
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template<class Elem>
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void Push(Stack<Elem> &sta,Elem item)
{
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sta.p[sta.top++]=item;
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}
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template<class Elem>
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void Pop(Stack<Elem> &sta,Elem &e)
{
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e=sta.p[--sta.top];
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}
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template<class Elem>
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bool IsEmpty(Stack<Elem> &sta)
{
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return sta.top==0;
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}
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template<class Elem>
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bool IsFull(Stack<Elem> &sta)
{
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return sta.top==sta.size;
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}
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//自定义队列
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template<class Elem>
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struct MyQuene
{
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int front;
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int rear;
37
Elem *p;
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int size;
39
};
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template<class Elem>
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void Set_Qsize(MyQuene<Elem> &Q,int n)
{
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Q.size=n;
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Q.front=0;
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Q.rear=0;
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Q.p=new Elem[Q.size];
46
}
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template<class Elem>
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void AddQuene(MyQuene<Elem> &Q,Elem item)
{
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Q.p[Q.rear++]=item;
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if(Q.rear==Q.size)
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Q.rear=0;
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}
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template<class Elem>
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void DeleteQuene(MyQuene<Elem> &Q,Elem& e)
{
55
e=Q.p[Q.front++];
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if(Q.front==Q.size)
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Q.front=0;
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}
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template<class Elem>
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Elem GetTop(Stack<Elem> &sta)
{
61
return sta.p[sta.top-1];
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}
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template<class Elem>
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bool IsEmpty(MyQuene<Elem> &Q)
{
65
return Q.front==Q.rear;
66
}
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template<class Elem>
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bool IsFull(MyQuene<Elem> &Q)
{
69
int len=Q.front<Q.rear?Q.rear-Q.front:Q.rear-Q.front+Q.size;
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return len==Q.size-1;
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}
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//定义运算符的优先级
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int GetPrecedence(char a)
{
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switch(a)
{
75
case '#':return 0;
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case '+':
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case '-':return 1;
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case '*':
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case '/':return 2;
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case '^':return 3;
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default:break;
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}
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}
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template<class Elem>
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void Middle_Bhind(Stack<Elem> &st,MyQuene<Elem>&Mq,char*A,int n)
{//中缀表达式转为后缀表达式(自己的实验需求)
86
Set_Ssize(st,n);
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Set_Qsize(Mq,n+1);
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char tem;
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int i=0;
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Push(st,'#');
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do
{
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if((A[i]>='0'&&A[i]<='9')||(A[i]>='A'&&A[i]<='Z')||(A[i]>='a'&&A[i]<='z'))
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AddQuene(Mq,A[i]);
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else if(A[i]=='(')
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Push(st,A[i]);
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else if(A[i]==')')
{
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while(GetTop(st)!='(')
{
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Pop(st,tem);
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AddQuene(Mq,tem);
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}
101
Pop(st,tem);
102
}
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else if(GetTop(st)=='(')
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Push(st,A[i]);
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else
{
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while(GetPrecedence(A[i])<=GetPrecedence(GetTop(st)))
{
107
Pop(st,tem);
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AddQuene(Mq,tem);
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}
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Push(st,A[i]);
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}
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i++;
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}while(A[i]!='\0');
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while(GetTop(st)!='#')
{
115
Pop(st,tem);
116
AddQuene(Mq,tem);
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}
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while(!IsEmpty(Mq))
{
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DeleteQuene(Mq,tem);
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printf("%c",tem);
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}
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putchar('\n');
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}
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void main()
{
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char str[Max];
126
Stack<char> st;
127
MyQuene<char>Mq;
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printf("请输入中缀表达式:");
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gets(str);
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printf("后缀表达式:");
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Middle_Bhind(st,Mq,str,Max);
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}
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posted on 2011-01-19 15:51
あ维wêiセ 阅读(6022)
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