中缀表达式转后缀表达式

来源：互联网发布：阿里云的天池数据赛编辑：程序博客网时间：2024/05/16 05:11

#include <stdio.h>#include <stdlib.h>#define ElementType chartypedef struct Node *PtrToNode;typedef PtrToNode Stack;typedef struct Node{    ElementType Element;    PtrToNode Next;};int IsEmpty(Stack S);Stack CreateStack();void DisposeStack(Stack S);void MakeEmpty(Stack S);void Push(ElementType X,Stack S);ElementType Top(Stack S);void Pop(Stack S);//判断栈是否为空int IsEmpty(Stack S){    return S->Next == NULL;}//创建链栈Stack CreateStack(){    Stack S = malloc(sizeof(struct Node));    if(S == NULL)    {        printf("No enough memory!");        return NULL;    }    S->Next = NULL;    MakeEmpty(S);    return S;}//清空栈void MakeEmpty(Stack S){    if(S == NULL)    {        printf("Use CreateStack First!");    }    else    {        while(!IsEmpty(S))        {            Pop(S);        }    }}//进栈void Push(ElementType X,Stack S){    PtrToNode Tmp;    Tmp = malloc(sizeof(struct Node));    if(Tmp != NULL)    {        Tmp->Element = X;        Tmp->Next = S->Next;        S->Next = Tmp;    }    else    {        printf("Out of space!");    }}//出栈void Pop(Stack S){    if(IsEmpty(S))    {        printf("The Stack is Empty!");    }    else    {        PtrToNode Tmp = S->Next;        S->Next = Tmp->Next;        free(Tmp);    }}//返回栈顶元素ElementType Top(Stack S){    if(IsEmpty(S))    {        printf("The stack is empty!");        return 0;    }    else    {        return S->Next->Element;    }}/******************************************************************根据字符该字符是否在栈中，返回该字符的优先级。*这里只处理+、-、*、/、(、)这些符号。*需要注意的是：如果(在栈中，它的优先级是最低的，不在栈中则是最高的*@param c:需要判断的字符*@param flag:字符是否在栈中，0表示在栈中，1表示不在栈中*****************************************************************/int GetPrecedence(char c,int flag){    if(c=='+' || c=='-')    {        return 1;    }    else if(c=='*' || c=='/')    {        return 2;    }    else if(c=='(' && flag==0)    {        return 0;    }    else if(c=='(' && flag==1)    {        return 3;    }    else    {        fprintf(stderr,"Input char is invalid!\n");        return -1;    }}/*****************************************************************判断一个字符是不是运算符*如果是合法的运算符+、-、*、/、(、)则返回0，否则返回1****************************************************************/int IsOperator(char c){    if(c=='+'||c=='-'||c=='*'||c=='/'||c=='('||c==')')    {        return 0;    }    else    {        return 1;    }}//中缀表达式转成后缀表达式char* InfixToPostfix(int i,char *ch,Stack S){    char * Output=malloc(sizeof(char)*i);//存放转换后的数组    int index=0;    char c;    while((c=*ch) != '\0')    {        //不是运算符，将该字符放进输出字符数组中。        if(IsOperator(c)==1)        {            Output[index++] = c;            ch++;        }        //是运算符        else        {            //如果此时栈为空，运算符进栈            if(IsEmpty(S))            {                Push(c,S);                ch++;                continue;            }            else            {                if(c==')')                {                    while(!IsEmpty(S) && Top(S) != '(')                    {                        Output[index++] = Top(S);                        Pop(S);                    }                    Pop(S);                    ch++;                    continue;                }                else                {                    int outPrecedence = GetPrecedence(c,1);                    while(!IsEmpty(S) && GetPrecedence(Top(S),0) >= outPrecedence)                    {                        Output[index++] = Top(S);                        Pop(S);                    }                    Push(c,S);                    ch++;                    continue;                }            }        }    }    while(!IsEmpty(S))    {        Output[index++] = Top(S);        Pop(S);    }    Output[index] = '\0';    return Output;}int main(void){    Stack S = CreateStack();    char *charSequence = "1+2*3+(4*5+6)*7";    char tmp;    char *out = InfixToPostfix(50,charSequence,S);    while((tmp=*out)!='\0')    {        printf("%c ",tmp);        out++;    }    printf("\n");    return 0;}

</pre><pre name="code" class="cpp">

运行结果：

<img src="http://img.blog.csdn.net/20150425150841363?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvZnVqaWFtaW5nMTIz/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt="" />

0 0