## Monday, November 30, 2015

### LeetCode OJ - Basic Calculator

Problem:

Please find the problem here.

Solution:

For the purpose of this problem, I wrote a simple top down recursive parser for this grammar.

Expression = Value | Expression + Value | Expression - Value
Value = Number | (Expression)

Note the grammar has left recursion which mean top-down recursive parsing will run into infinite loop, eventually run out of stack, the innovative part of this solution, unlike traditional parsing, is to parse the string backwards, if we do that, then there is no right recursion, and now we are fine.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/basic-calculator/

#include "LEET_BASIC_CALCULATOR.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_BASIC_CALCULATOR
{
class Solution
{
public:
// Scanner states
int position;
char token;
string* expression;

// Expression = Value
// Expression = Expression + Value
// Expression = Expression - Value
// Value = Number
// Value = ( Expression )

// The key innovation in this solution is parsing backward and thus eliminate the left recursion problem of the left associative grammar
void scan()
{
while (true)
{
if (position >= 0)
{
token = (*expression)[position--];
}
else
{
token = '\0';
}
if (token != ' ')
{
break;
}
}
}

bool ParseValue(int* result)
{
if (token >= '0' && token <= '9')
{
return ParseNumber(result);
}
else if (token == ')')
{
scan();
int bracketedExpression;
if (ParseExpression(&bracketedExpression))
{
*result = bracketedExpression;
if (token == '(')
{
scan();
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}
else
{
return false;
}
}

bool ParseNumber(int* result)
{
int base = 1;
*result = 0;
while (true)
{
if (token >= '0' && token <= '9')
{
int digit = token - '0';
*result = *result + base * digit;
base *= 10;
scan();
}
else
{
return true;
}
}
return false;
}

bool ParseExpression(int* result)
{
if ((token >= '0' && token <= '9') || token == ')')
{
int tail;
if (ParseValue(&tail))
{
if (token == '+')
{
scan();
{
*result = head + tail;
return true;
}
}
else if (token == '-')
{
scan();
{
*result = head - tail;
return true;
}
}
else if (token == '\0' || token == '(')
{
*result = tail;
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}

return false;
}

int calculate(string s)
{
position = s.size() - 1;
expression = &s;
scan();
int result;
if (ParseExpression(&result))
{
return result;
}
cout << "Parse error!" << endl;
return 0;
}
};
};

using namespace _LEET_BASIC_CALCULATOR;

int LEET_BASIC_CALCULATOR()
{
Solution solution;
cout << (solution.calculate("1 + 1") == 2) << endl;
cout << (solution.calculate(" 2-1 + 2 ") == 3) << endl;
cout << (solution.calculate("(1+(4+5+2)-3)+(6+8)") == 23) << endl;
return 0;
}