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]
Analysis:
Divide and conquer
class Solution { void calculate(vector<int>&res, vector<int>&l, vector<int>&r, char op) { for(auto e:l) { for(auto m:r) { if(op == '+') res.push_back(e+m); else if(op == '-') res.push_back(e-m); else if(op == '*') res.push_back(e*m); else if(op == '/') { if(m!=0)res.push_back(e/m); else res.push_back(e>0?INT_MAX:INT_MIN); } } } } void solve(vector<int> &res, const string & input) { bool found = false; for(int i=1;i<input.length();i++) { if(input[i] == '+' || input[i] == '-' || input[i] == '*' || input[i] == '/') { vector<int> l, r; solve(l, input.substr(0,i)); solve(r, input.substr(i+1)); if(!found) found = true; calculate(res, l, r, input[i]); } } if(!found) res.push_back(stoi(input)); } public: vector<int> diffWaysToCompute(string input) { vector<int> res; solve(res, input); return res; } };