Different Ways to Add Parentheses

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) = 10Output: 

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

题目： 判断有多少种可能的运算结果。算法：递归解决问题，相当于全排列问题。每次都以符号分割两个子串，然后计算左边和右边的运算结果放在result里。代码#include <iostream>
#include <string>
#include <vector>
using namespace std;
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int>result;
if(input.size()==0)
return result;
for (int i = 0; i < input.size(); ++i) {
if(ispunct(input[i])){
vector<int>left = diffWaysToCompute(input.substr(0,i));
vector<int>right = diffWaysToCompute(input.substr(i+1));
for (int j = 0; j < left.size(); ++j) {
for (int k = 0; k < right.size(); ++k) {
switch (input[i]){
case '+':
result.push_back(left[j]+right[k]);
break;
case '-':
result.push_back(left[j]-right[k]);
break;
case '*':
result.push_back(left[j]*right[k]);
break;
default:
break;

}
}
}
}
}
if(result.empty()){
result.push_back(stoi(input));
}
return result;

}
};
int main() {
Solution s;
vector<int> r = s.diffWaysToCompute("2*3-4*5");
cout<<1;
return 0;
}