Different Ways to Add Parentheses

Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are

+

,

-

and

*

.

Example 1

Input:

"2-1-1"

.

((2-1)-1) = 0
(2-(1-1)) = 2

Output:

[0, 2]

Example 2

Input:

"2*3-4*5"

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

Output:

[-34, -14, -10, -10, 10]

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> vals;
vector<char> ops;
parse(input,vals,ops);
if (vals.size() == 1)
{
return vals;
}
return diffWaysToCompute(vals, ops, 0, vals.size());
}
private:
vector<int> diffWaysToCompute(vector<int> const &vals, vector<char> const & ops, int beg, int end)
{
vector<int> output;
int size = end - 1;
for (size_t i = beg + 1; i <= size; i++)
{
vector<int> left = diffWaysToCompute(vals, ops, beg, i);
vector<int> right = diffWaysToCompute(vals, ops, i, end);
for (int il = 0; il < left.size(); ++il)
{
for (int ir = 0; ir < right.size(); ++ir)
{
output.push_back(calc(left[il], right[ir],ops[i - 1]));
}
}
}
if(output.empty())
{
output.push_back(vals[beg]);
}
return output;
}
void parse(string & input,vector<int> &vals,vector<char> &ops)
{
auto  it = input.begin();
auto const  itEnd = input.end();
while (it != itEnd)
{
char ch = *it;
if (isdigit(ch))
{
vals.push_back(getVal(it, itEnd));
}
else
{
ops.push_back(ch);
++it;
}
}
}
int calc(int a, int b, char op)
{
switch(op)
{
case '+': return a + b;
case '-': return a - b;
case '*': return a * b;
}
return  1;
}
int getVal(string::iterator & it, string::iterator const &itEnd)
{
int val = 0;
while (it != itEnd && isdigit(*it))
{
val = val * 10 + *it - '0';
++it;
}
return val;
}
};