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]

Credits:

Special thanks to  for adding this problem and creating all test cases.

 

与思路类似，可以对照看。

本题参考的做法

左右子串分别计算所有可能，然后全排列。

class Solution {public:    vector
     
       diffWaysToCompute(string input) {        vector
      
        ret;        for(int i = 0; i < input.size(); i ++)        {            if(input[i] == '+' || input[i] == '-' || input[i] == '*')            {                vector
       
         left = diffWaysToCompute(input.substr(0, i));                vector
        
          right = diffWaysToCompute(input.substr(i+1));                for(int j = 0; j < left.size(); j ++)                {                    for(int k = 0; k < right.size(); k ++)                    {                        if(input[i] == '+')                            ret.push_back(left[j] + right[k]);                        else if(input[i] == '-')                            ret.push_back(left[j] - right[k]);                        else                            ret.push_back(left[j] * right[k]);                    }                }            }        }        if(ret.empty())            ret.push_back(atoi(input.c_str()));        return ret;    }};