逆波兰计数器 栈的实现

又称后缀计数器

用栈把原式写为逆波兰式子，保存在数组里面并打印。

用栈根据逆波兰式子的形式计算出结果。

以下代码：

#include
#include
#include
#define INITSIZE 20	
#define INCREASE 5
typedef struct Stack {
	char*top;
	char*base;
	int stackSize;
}Stack;
void InitStack(Stack*s) {
	s->base = (char*)malloc(sizeof(char));//相当于为首地址申请空间
	s->top = s->base;
	s->stackSize = INITSIZE;
}
void Push(Stack*s, char c) {
	if (s->top- s->base>=s->stackSize) {
		s->base = (char*)realloc(s->base, (s->stackSize + INCREASE) * sizeof(char));
		//为s->base赋一个新的数值空间，并把s->base移过去
		s->top = s->base + s->stackSize;//把s->top
		s->stackSize = s->stackSize + INCREASE;
	}
	*(s->top++) = c;
}
void Pop(Stack*s, char*e) {
	if (s->top == s->base)
		return;
	*e = *(--s->top);
}
int getLen(Stack*s) {
	return s->top - s->base;
}
void ClearStack(Stack*s) {
	s->top = s->base;
}

//实现逆波兰计数器（后缀计数器）
char b[100] = { 0 };
typedef struct stack2 {
	double base[20];
	int top;
}Stack2;
int NiBoLan(Stack*s, char*a) {
	int i =0, j = 0;
	char e;
	while (a[i]) {
		if (a[i] >= '0'&&a[i] <= '9')
			b[j++] = a[i];
		else if (a[i] == ')') {
			Pop(s, &e);//s已经是个指针了，不能再引用了
			while (e != '(') {
				b[j++] = ' ';//把逆波兰实现的数据存到数组b上，数字和符号之间以空格相隔
				b[j++] = e;
				Pop(s, &e);
			}
		}
		else if (a[i] == '+' || a[i] == '-') {
			b[j++] = ' ';
			if (s->top == s->base || *(s->top - 1) == '(')
				Push(s, a[i]);
			else{//注意else if的区别
				while (getLen(s) && *(s->top - 1) != '(') {
					Pop(s, &e);
					b[j++] = e;
					b[j++] = ' ';
				}
				Push(s, a[i]);
			}
		}
		else if (a[i] == '*' || a[i] == '/' || a[i] == '(') {
			if(a[i]!='(')
				b[j++] = ' ';
			Push(s, a[i]);
		}
		else return 0;
		i++;
	}
	while (s->top!=s->base) {
		Pop(s, &e);
		b[j++] = ' ';
		b[j++] = e;
	}
	return 1;
}
double result(Stack2 s) {
	int sum = 0, i = 0,j;
	char num[10];
	double e, d;
	while (b[i]) {
		j = 0;
		if (b[i] >= '0'&&b[i] <= '9') {
			while (b[i] != ' '&&b[i] >= '0'&&b[i] <= '9') {
				num[j++] = b[i];
				num[j] = 0;//要在num中字符的后一位加上'\0'，表示字符串的终止
				i++;
			}
			e = atof(num);
			s.base[++s.top] = e;//进行入栈
		}
		if (s.top > 0) {
			switch (b[i])
			{
			case '+':
				s.base[s.top - 1] = s.base[s.top] + s.base[s.top - 1];
				s.top--;
				break;
			case '-':
				s.base[s.top - 1] = s.base[s.top - 1] - s.base[s.top];
				s.top--;
				break;
			case '*':
				s.base[s.top - 1] = s.base[s.top] * s.base[s.top - 1];
				s.top--;
				break;
			case '/':
				s.base[s.top - 1] = s.base[s.top - 1] / s.base[s.top];
				if (s.base[s.top] == 0) {
					printf("出现浮点错误！\n");
					return -1;
				}
				s.top--;
				break;
			default:
				break;
			}
		}
		i++;
	}
	return s.base[s.top];
}

int main() {
	Stack s;
	InitStack(&s);
	Stack2 s2 = {
		{ 0 },-1
	};//对s2赋予初值
	char a[100] = { 0 };
	printf("请输入一个算式：");
	scanf("%s", a);
	int IsCorrect = NiBoLan(&s, a);
	if (IsCorrect) {
		for (int i = 0; b[i]; i++)
			printf("%c", b[i]);
		printf("\n");
	}
	else printf("输入的算式有误！\n");
	printf("最后结果为：%.2lf\n", result(s2));
	return 0;
}