【LeetCode】241. Different Ways to Add Parentheses

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 @mithmatt for adding this problem and creating all test cases.

与Unique Binary Search Trees II思路类似，可以对照看。

本题参考Gcdofree的做法

左右子串分别计算所有可能，然后全排列。

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> ret;
        for(int i = ; i < input.size(); i ++)
        {
            if(input[i] == '+' || input[i] == '-' || input[i] == '*')
            {
                vector<int> left = diffWaysToCompute(input.substr(, i));
                vector<int> right = diffWaysToCompute(input.substr(i+));
                for(int j = ; j < left.size(); j ++)
                {
                    for(int k = ; 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;
    }
};