241. 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]

给多项式加括号，输出各种结果。

以运算符为基准分割字符串，字符串总的输出结果就等于 左子串结果op右子串结果。

为了加快效率，可以为每个子串记录运算结果，防止重复计算。

class Solution {
private:
    unordered_map<string, vector<int>> result;
public:
    vector<int> diffWaysToCompute(string input) {
        unsigned int len = input.size();
        unsigned int i = 0;
        vector<int> res;
        for (; i<len; ++i) {
            char c = input[i];
            if (c == '+' || c == '-' || c == '*') {
                string s1 = input.substr(0, i);
                vector<int> v1;
                v1 = result.find(s1) == result.end()? diffWaysToCompute(s1): result[s1];
                string s2 = input.substr(i+1);
                vector<int> v2;
                v2 = result.find(s2) == result.end()? diffWaysToCompute(s2): result[s2];
                for (auto &n1: v1)
                    for (auto &n2: v2) {
                        if (c == '+')
                            res.push_back(n1 + n2);
                        else if (c == '-')
                            res.push_back(n1 - n2);
                        else
                            res.push_back(n1 * n2);
                    }
            }
        }
        if (res.empty())
            res.push_back(stoi(input));
        result[input] = res;
        return res;
    }
};