【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;

    }

};

巴特西

【LeetCode】241. Different Ways to Add Parentheses

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