Different Ways to Add Parentheses -- leetcode

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]




基本思路：

以每個操作符作爲分隔符，將表達式分成左右兩個新的子表達式。

對兩個子表達式進行遞歸求解。然後對返回值進行笛卡爾積。

特例： 當找不到操作符時，只須將表達式轉換成整數。

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> ans;
        for (int i=0; i<input.size(); i++) {
            if (!isdigit(input[i])) {
                auto left = diffWaysToCompute(input.substr(0, i));
                auto right = diffWaysToCompute(input.substr(i+1));
                for (auto l: left) {
                    for (auto r: right)
                        ans.push_back(input[i]=='+' ? l+r : (input[i]=='-' ? l-r : l*r));
                }
            }
        }
        if (ans.empty())
            ans.push_back(stoi(input));
        return ans;
    }
};