表达式问题-C

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wx6292bf1d80fe8 2023-01-17 07:20:16 博主文章分类：C++算法题 ©著作权

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表达式转后缀表达式：

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h> 
#include<assert.h>
#define INITSIZE  20
#define INCREMENT 10
#define MAXBUFFER 20
#define LEN  sizeof(Elemtype)
/*栈的动态分配存储结构*/ 
typedef char Elemtype;
typedef struct{
  Elemtype *base;
  Elemtype *top;
  int StackSize;
}SqStack;

/*初始化栈*/
void InitStack(SqStack *S) {
  S->base=(Elemtype*)malloc(LEN*INITSIZE);
  assert(S->base !=NULL);
  S->top=S->base;
  S->StackSize=INITSIZE;
}

/*压栈操作*/ 
void PushStack(SqStack *S,Elemtype c) {
  if(S->top - S->base >= S->StackSize) {
    S->base=(Elemtype*)realloc(S->base,LEN*(S->StackSize+INCREMENT));
    assert(S->base !=NULL);
    S->top =S->base+S->StackSize;
    S->StackSize+=INCREMENT;
  }
  *S->top++ = c;
}
/*求栈长*/
int StackLength(SqStack *S) {
  return (S->top - S->base);
}
/*弹栈操作*/
int PopStack(SqStack *S,Elemtype *c) {
  if(!StackLength(S)) {
    return 0;
  }
  *c=*--S->top;
  return 1;
}

/*中缀转后缀函数*/
void Change(SqStack *S,Elemtype str[]) {
  int i=0;
  Elemtype e;
  InitStack(S);
  while(str[i]!='\0') {
    while(isdigit(str[i])) {
      /*过滤数字字符，直接输出，直到下一位不是数字字符打印空格跳出循环 */
      printf("%c",str[i++]);
      if(!isdigit(str[i])) {
        printf(" ");
      }
    }
    if(str[i]=='+'||str[i]=='-') {
      if(!StackLength(S)) {
        PushStack(S,str[i]);
      } else {
        do {
          PopStack(S,&e);
          if(e=='(')
          {
            PushStack(S,e);
          } else {
            printf("%c ",e);
          }
        } while( StackLength(S) && e != '(' ); 
        PushStack(S,str[i]);
      }
    }
    /*当遇到右括号是，把括号里剩余的运算符弹出，直到匹配到左括号为止
    左括号只弹出不打印（右括号也不压栈）*/
    else if(str[i]==')') {
      PopStack(S,&e);
      while(e!='(') {
        printf("%c ",e);
        PopStack(S,&e);
      }
    } else if(str[i]=='*'||str[i]=='/'||str[i]=='(') {
      /*乘、除、左括号都是优先级高的，直接压栈*/
      PushStack(S,str[i]);
    } else if(str[i]=='\0') {
      break;
    } else {
      printf("\n输入格式错误！\n");
      return ;
    }
    i++;
  }
  /*最后把栈中剩余的运算符依次弹栈打印*/
  while(StackLength(S)) {
    PopStack(S,&e);
    printf("%c ",e);
  }
}
int main() {
  printf("请输入表达式,如：(2+(((2+6)*6+2)*2+6)*2):\n\n"); 
  Elemtype str[MAXBUFFER];
  SqStack S;
  gets(str);
  printf("\n转化为后缀表达式：");
  Change(&S,str);
  return 0;
}

后缀表达式的计算：

#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
#include<ctype.h>
#define INITSIZE  20
#define INCREMENT 10
#define MAXBUFFER 10
#define LEN   sizeof(Elemtype)
/*栈的动态分配顺序存储结构*/
typedef double Elemtype;

//结构体 
typedef struct{
  Elemtype *base;
  Elemtype *top;
  int StackSize; 
}SqStack;
//初始化栈 
void InitStack(SqStack *S) {
  S->base=(Elemtype*)malloc(LEN*INITSIZE);
  assert(S->base != NULL);
  S->top=S->base;
  S->StackSize=INITSIZE;
}
//压栈 
void PushStack(SqStack *S,Elemtype e) {
  if(S->top - S->base >= S->StackSize)
  {
    S->base=(Elemtype*)realloc(S->base,(S->StackSize+INCREMENT)*LEN);
    assert(S->base !=NULL);
    S->top=S->base+S->StackSize;
    S->StackSize+=INCREMENT;
  }
  *S->top =e;
   S->top++;
} 
//出栈 
void PopStack(SqStack *S,Elemtype *e) {
  *e=*--S->top;
} 
//计算-关键函数 
void CalFunction(SqStack *S,char str[]) {
  Elemtype number,e,d;
  char arr[MAXBUFFER];
  int i=0,j=0; 
  InitStack(S);
  while(str[i]!='\0') {
    while(isdigit(str[i])||str[i]=='.') { //过滤数字
      arr[j++]=str[i++];
      arr[j]='\0';
      if( j >= MAXBUFFER ) {
        printf("输入单个数据过大！\n");
        return ;
      }
      if(str[i]==' ') {
        number=atof(arr);    //利用atof函数将数字字符转化为double型数据
        PushStack(S,number); //将转换的数进行压栈
        j=0;
        break;
      }
    }
    switch(str[i]) {
      case '+':
        PopStack(S,&e);
        PopStack(S,&d);
        PushStack(S,d+e);
        break;
      case '-':
        PopStack(S,&e);
        PopStack(S,&d);
        PushStack(S,d-e);
        break;
      case '*':
        PopStack(S,&e);
        PopStack(S,&d);
        PushStack(S,d*e);
        break;
      case '/':
        PopStack(S,&e);
        PopStack(S,&d);
        if(e == 0) {
          printf("输入出错，分母为零！\n");
          return ;
        }
        PushStack(S,d/e);
        break;
    }
    i++;  
  }
  PopStack(S,&e);
  printf("计算结果为：%lf",e);  
}
//主函数 
int main() {
  char str[100];
  SqStack S;
  printf("请按逆波兰表达式输入数据,每个数据之间用空格隔开,如：2 2 6 + 6 * 2 + 2 * 6 + 2 * + \n");
  gets(str);
  CalFunction(&S,str);
  printf("\n\n");
  return 0;
}