[LeetCode] 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.

先想到是就是递归了，下面是AC代码。可能有更好的解法，想到了再回来补充。

class Solution {
public:
    int compute(int a, int b, char op) {
        switch (op) {
            case '+': return a + b;
            case '-': return a - b;
            case '*': return a * b;
        }
        return 1;
    }
    vector<int> diffWaysToCompute(string input) {
        int val = 0, idx = 0;
        while (idx < input.length() && isdigit(input[idx])) {
            val *= 10;
            val += input[idx++] - '0';
        }
        if (idx == input.length()) return {val};
        vector<int> res;
        vector<int> left, right;
        for (int i = 0; i < input.length(); ++i) {
            if (!isdigit(input[i])) {
                left = diffWaysToCompute(input.substr(0, i));
                right = diffWaysToCompute(input.substr(i + 1, input.length() -1 - i));
                for (int j = 0; j < left.size(); ++j) {
                    for (int k = 0; k < right.size(); ++k) {
                        res.push_back(compute(left[j], right[k], input[i]));
                    }
                }
            }
        }
        return res;
    }
};