C#中，中缀表达式如何转化为后缀表达式？

用C#代码写出转换流程，而不是给出一个四则运算符后用大脑去转换的，谢谢各位大侠的帮助

答案：这是算法
#include<stdio.h> 
#include<stdlib.h> 
#define TRUE 1 
#define FALSE 0 
#define MAXNUM 100 
typedef int DataType; 
typedef  struct 
{ 
DataType s[MAXNUM]; 
int t; 
}SeqStack,*PSeqStack;


 //构造一个空栈
PSeqStack createEmptyStack_seq() 
{
 PSeqStack pastack; 
 pastack=(SeqStack*)malloc(sizeof( SeqStack));
 
 if (pastack==NULL) 
 { 
  printf("空间不够!!\n"); 
 } 
 else 
 { 
  pastack->t=-1; 
 } 
 return pastack;//返回空栈顶 
} 


//清空栈
int isEmptyStack_seq(PSeqStack pastack) 
{ 
 return pastack->t==-1; 

}


 //入栈
void push_seq(PSeqStack pastack, DataType x) 
{ 
 if (pastack->t >= MAXNUM - 1) 

    printf("上溢!\n"); 
 else 
 { 
  pastack->t = pastack->t+1; 
        pastack->s[pastack->t]=x; 
 } 
} 

//出栈
void pop_seq(PSeqStack pastack) 
{ 
 if (pastack->t==-1) 
 { 
  printf("下溢!\n"); 
 } 
 else 
 { 
  pastack->t = pastack->t-1; 
 } 
} 

//返回栈顶元素的值
DataType top_seq(PSeqStack pastack) 
{ 
return pastack->s[pastack->t]; 
} 



/*将中缀表达式转换为后缀表达式，顺利转换返回true，若转换过程中发现中缀表达式非法则返回false*/ 
int infixtoSuffix(const char* infix, char* suffix) 
{ 
 int state_int = FALSE; /*state_int记录状态，等于true表示刚读入的是数字字符，等于false表示刚读入的是运算符， 
                           设置这个变量是为了在每输出一个整数后输出一个空格，以免连续输出的两个整数混在一起。*/ 
 char c, c2; 
    PSeqStack ps = createEmptyStack_seq(); /*构造一个运算符栈*/ 
    int i, j = 0; 
    if(infix[0]=='\0') 
 { 
        return FALSE; /*不允许出现空表达式*/ 
 } 
    for(i=0; infix[i]!='\0';i++) 
 { 
        c=infix[i]; 

        switch(c) 
  {
  case ' ': 
        case '\t': 
        case '\n': 
   if(state_int== TRUE) 
   { 
    suffix[j++]=' ';/*状态从true转换为false时输出一个空格*/ 
   } 
   state_int= FALSE; 
   break; /*遇到空格或制表符忽略*/ 
  case '0': 
        case '1': 
        case '2':
  case '3': 
  case '4': 
  case '5':
  case '6': 
  case '7':
  case '8': 
  case '9':
   state_int =TRUE; 
            suffix[j++]=c; /*遇到数字输出*/ 
   break; 
        case '(': 
   if(state_int==TRUE) 
   { 
    suffix[j++]=' ';/*状态从true转换为false时输出一个空格*/ 
   } 
   state_int=FALSE; 

            push_seq(ps, c); /*遇到左括号，入栈*/ 
   break; 
  case ')': 
    if(state_int== TRUE) 
   { 
    suffix[j++]=' ';/*状态从true转换为false时输出一个空格*/ 
   } 
   state_int=FALSE; 
   c2 = ')'; 
   while(!isEmptyStack_seq(ps)) 
   { 
    c2=top_seq(ps);/*取栈顶*/ 
    pop_seq(ps); /*出栈*/ 
    if(c2 =='(') 
    { 
     break; 
    } 
    suffix[j++]=c2; 
   } 
   if(c2 !='(') 
   { 
    free(ps); 
    suffix[j++]='\0'; 
    return FALSE; 
   } 
   break; 
  case '+': 
  case '-': 
   if(state_int == TRUE) 
   { 
    suffix[j++] = ' '; 
   } 
            state_int = FALSE; 
   while(!isEmptyStack_seq(ps)) 
   { 
    c2 = top_seq(ps); 
               if(c2 =='+'|| c2 =='-'|| c2 == '*' || c2 == '/') 
      { 
                 pop_seq(ps); 
                 suffix[j++] = c2; 
      } 
      else if(c2=='(') 
      { 
                 break; 
      } 
   } 
   push_seq(ps, c); 
   break; 
  case '*': 
        case '/': 
   if(state_int==TRUE) 
   { 
    suffix[j++] = ' '; 
   } 
   state_int = FALSE; 
   while(!isEmptyStack_seq(ps)) 
   { 
    c2 = top_seq(ps); 
    if(c2 == '*' || c2 == '/') 
    { 
     pop_seq(ps); 
     suffix[j++] = c2; 
    } 
    else if(c2=='+'||c2=='-'||c2=='(') 
    { 
     break; 
    } 
   } 
   push_seq(ps, c); 
   break; 
  default: 
   free(ps); 
   suffix[j++] = '\0'; 
            return FALSE; 
} 
} 
if(state_int == TRUE) 
{ 
   suffix[j++] = ' '; 
} 
while(!isEmptyStack_seq(ps)) 
{ 
c2 = top_seq(ps); 
pop_seq(ps); 
if(c2 == '(') 
{ 
free(ps); 
suffix[j++] = '\0'; 
return FALSE; 
} 
suffix[j++] = c2; 
} 
free(ps); 
suffix[j++] = '\0'; 
return TRUE; 
} 


//计算后缀
int calculateSuffix(const char * suffix, int * presult) 
{ 

int state_int = FALSE; 
PSeqStack ps = createEmptyStack_seq(); 
int num = 0, num1, num2; 
int i; 
char c; 
for(i = 0; suffix[i] != '\0'; i++) 
{ 
c = suffix[i]; 
switch(c) 
{ 
case '0': 
case '1': 
case '2': 
case '3': 
case '4': 
case '5': 
case '6': 
case '7': 
case '8': 
case '9': 
if(state_int == TRUE) 
{ 
num = num * 10 + c - '0'; 
} 
else 
{ 
num = c - '0'; 
} 
state_int = TRUE; 
break; 
case ' ': 
case'\t': 
case '\n': 
if (state_int == TRUE) 
{ 
push_seq(ps, num); 
state_int = FALSE; 
} 
break; 
case '+': 
case '-': 
case '*': 
case '/': 
if(state_int == TRUE) 
{ 
push_seq(ps, num); 
state_int = FALSE; 
} 
if(isEmptyStack_seq(ps)) 
{ 
free(ps); 
return FALSE; 
} 
num2 = top_seq(ps); 
pop_seq(ps); 
if(isEmptyStack_seq(ps)) 
{ 
free(ps); 
return FALSE; 
} 
num1 = top_seq(ps); 
pop_seq(ps); 
if(c == '+') 
{ 
push_seq(ps, num1 + num2); 
} 
if(c == '-') 
{ 
push_seq(ps, num1 - num2); 
} 
if(c == '*') 
{ 
push_seq(ps, num1 * num2); 
} 
if(c == '/') 
{ 
push_seq(ps, num1 / num2); 
} 
break; 
default: 
free(ps); 
return FALSE; 
} 
} 
*presult = top_seq(ps); 
pop_seq(ps); 
if(!isEmptyStack_seq(ps)) 
{ 
free(ps); 
return FALSE; 
} 
free(ps); 
return TRUE; 
} 
void getline(char* line, int limit) 
{ 
char c; 
int i=0; 
while(i<limit-1&&(c = getchar())!=EOF&&c!='\n') 
{ 
line[i++]=c; 
} 
line[i]='\0'; 
} 


//主函数
void main() 
{ 
char c, infix[MAXNUM], suffix[MAXNUM]; 
int result; 
int flag = TRUE; 
while(flag == TRUE) 
{ 
printf("请输入任意一个整数算术表达式:\n"); 
getline(infix, MAXNUM); 
if(infixtoSuffix(infix, suffix) == TRUE) 
{ 
printf("所有后缀为:%s\n", suffix); 
} 
else 
{ 
printf("无效缀!\n"); 
printf("\n继续? (y/n)"); 
scanf("%c", &c); 
if(c == 'n' || c == 'N') 
{ 
flag = FALSE; 
} 
while(getchar() != '\n'); 
{ 
printf("\n"); 
} 
continue; 
} 

if(calculateSuffix(suffix, &result) == TRUE) 
{ 
printf("结果为:%d\n", result); 
} 
else 
{ 
printf("非法后缀!\n"); 
} 
printf("\n继续? (y/n)"); 
scanf("%c", &c); 
if(c == 'n' || c == 'N') 
{ 
flag = FALSE; 
} 
while(getchar() != '\n'); 
{ 
printf("\n"); 
} 
} 
} 

上一个：C#问题 各位大哥大姐们请帮帮忙啊！！！！！！
下一个：C#winform中如何把表导出到EXCEL