241. 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> result; int size = input.size(); for (int i = 0; i < size; i++) { char cur = input[i]; if (cur == '+' || cur == '-' || cur == '*') { // Split input string into two parts and solve them recursively vector<int> result1 = diffWaysToCompute(input.substr(0, i)); vector<int> result2 = diffWaysToCompute(input.substr(i+1)); for (auto n1 : result1) { for (auto n2 : result2) { if (cur == '+') result.push_back(n1 + n2); else if (cur == '-') result.push_back(n1 - n2); else result.push_back(n1 * n2); } } } } // if the input string contains only number if (result.empty()) result.push_back(atoi(input.c_str())); return result; } };