Leetcode oj:different Ways to add parentheses (method of adding parentheses in different locations)

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 ((1-1)) = 2

Output:[0, 2]

Example 2

Input:"2*3-4*5"

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

Output:[-34, -14, -10, -10, 10]

The main idea here is that the left and right sub-strings are calculated separately, then do the whole arrangement on the line, there may be a very repetitive calculation so that the efficiency is lower

1 classSolution {2  Public:3vector<int> Diffwaystocompute (stringinput) {4vector<int>result;5         intLength =input.length ();6          for(inti =0; i < length; ++i) {7             Charc =Input[i];8             if(c = ='+'|| c = ='-'|| c = ='*'){9                 stringInputleft = Input.substr (0, i);Ten                 stringInputright = Input.substr (i+1); Onevector<int>leftresult =Diffwaystocompute (inputleft); Avector<int>rightresult =Diffwaystocompute (inputright); -                  for(intj =0; J < Leftresult.size (); ++j) { -                      for(intK =0; K < Rightresult.size (); ++k) { the                         if(c = ='+') -Result.push_back (Leftresult[j] +rightresult[k]); -                         Else if(c = ='-') -Result.push_back (Leftresult[j]-rightresult[k]); +                         Else if(c = ='*') -Result.push_back (Leftresult[j] *rightresult[k]); +                     } A                 } at             }        -         } -         if(Result.empty ())//The main function of this step is to tell the final character of the divide to be converted into numbers - result.push_back (Stoi (input)); -         returnresult; -     } in};

Leetcode oj:different Ways to add parentheses (method of adding parentheses in different locations)