华为面试题：四则运算 C语言源码

四则运算

描述:

请实现如下接口

/* 功能：四则运算

* 输入：strExpression：字符串格式的算术表达式，如: "3+2*{1+2*[-4/(8-6)+7]}"

* 返回：算术表达式的计算结果

*/

public static int calculate(String strExpression)

{

/* 请实现*/

return 0;

}

约束：

pucExpression字符串中的有效字符包括[‘0’-‘9’],‘+’,‘-’, ‘*’,‘/’ ,‘(’， ‘)’,‘[’, ‘]’,‘{’ ,‘}’。

pucExpression算术表达式的有效性由调用者保证;

知识点:  栈

题目来源:  内部整理

练习阶段:  中级

运行时间限制: 10Sec

内存限制: 128MByte

输入:

输入一个算术表达式

输出:

得到计算结果

样例输入:

3+2*{1+2*[-4/(8-6)+7]}

样例输出:

25

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#define MAX_PATH 256
char stack_all[MAX_PATH] = {0};
char stack_symbol[MAX_PATH] = {0};
char stack_num[MAX_PATH] = {0};
int num_all=0;
int num_symbol =0;
int num_num = 0;

int cal(int a,int b,char op)
{
	switch(op)
	{
	case '+':
		return a+b;
	case '-':
		return a-b;
	case '*':
		return a*b;
	case '/':
		if (b==0)
		{
			return 0;
		}
		return a/b;
	default:
		return 0;
	}
}
//有效字符是0-9
int calculate(char *strExpression)
{

	int length = strlen(strExpression);
	for (int i=length-1;i>=0;i--)
	{
		char t = strExpression[i];
		if(t=='*' || t=='/' || t==')' || t==']' || t=='}')
		{
			stack_symbol[num_symbol] = t;
			num_symbol++;
		}
		else if (t=='(')
		{
			while(true)
			{
				if (stack_symbol[num_symbol-1]==')')
				{
					num_symbol--;
					break;
				}
				stack_all[num_all]=stack_symbol[num_symbol-1];
				num_all++;
				num_symbol--;
			}
		}
		else if (t=='[')
		{
			while(true)
			{
				if (stack_symbol[num_symbol-1]==']')
				{
					num_symbol--;
					break;
				}
				stack_all[num_all]=stack_symbol[num_symbol-1];
				num_all++;
				num_symbol--;
			}
		}
		else if (t=='{')
		{
			while(true)
			{
				if (stack_symbol[num_symbol-1]=='}')
				{
					num_symbol--;
					break;
				}
				stack_all[num_all]=stack_symbol[num_symbol-1];
				num_all++;
				num_symbol--;
			}
		}
		else if(t=='+' || t=='-')
		{
			while((stack_symbol[num_symbol-1]=='*' || stack_symbol[num_symbol-1]=='/') && num_symbol!=0 )
			{
				stack_all[num_all]=stack_symbol[num_symbol-1];
				num_all++;
				num_symbol--;
			}
			stack_symbol[num_symbol] = t;
			num_symbol++;
		}
		else
		{
			stack_all[num_all] = t;
			num_all++;
		}
	}
	for (int i=num_symbol-1;i>=0;i--)
	{
		stack_all[num_all]=stack_symbol[i];
		num_all++;
	}

	for (int i=0;i<num_all;i++)
	{
		char t = stack_all[i];
		if (t>='0'&&t<='9')
		{
			stack_num[num_num]=t-'0';
			num_num++;
		}
		else
		{
			stack_num[num_num-2] = cal(stack_num[num_num-1],stack_num[num_num-2],t);
			num_num--;
		}
	}
	return (int)stack_num[0];

} 

void init(char *buffer,char *buffer2)
{
	int length = strlen(buffer);
	int k = 0;
	for (int i=0;i<length;i++)
	{
		char t = buffer[i];
		if (t=='+'||t=='-')
		{
			if(i>=1 && (buffer[i-1]=='(' || buffer[i-1]=='[' || buffer[i-1]=='{'))
			{
				//可以不使用括号
// 				buffer2[k]='0';
// 				k++;
// 				buffer2[k]=buffer[i];
// 				k++;
				buffer2[k]='(';
				k++;
				buffer2[k]='0';
				k++;
				buffer2[k]=buffer[i];
				k++;
				buffer2[k]=buffer[i+1];
				i++;
				k++;
				buffer2[k]=')';
				k++;
			}
			else
			{
				buffer2[k]=buffer[i];
				k++;
			}
		}
		else
		{
			buffer2[k]=buffer[i];
			k++;
		}
	}
}
/*
2 3
4 6
8 10
特比注意：负数需要补0为0-正数。
*/
int main()
{
	int m=0,n=0;
	char buffer[MAX_PATH] = {0};
	gets(buffer);
//	char buffer[MAX_PATH] = "3+2*{1+2*[-4/(8-6)+7]}";
	char buffer2[MAX_PATH] = {0};
	init(buffer,buffer2);
// 	printf(buffer2);
// 	printf("\n");
	int ret = calculate(buffer2);
//	printf(stack_all);
//	printf("\n");
	printf("%d",ret);
	return 0;
}