Different Ways to Add Parentheses问题及解法

问题描述：

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 *.

示例：

Input: "2-1-1".

((2-1)-1) = 0(2-(1-1)) = 2

Output: [0, 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]

问题分析：

对与本题，我们可以利用分治法，将问题转换为各个子问题求解。

过程详见代码：

class Solution {public:    vector<int> diffWaysToCompute(string input) {vector<int> res;int size = input.length();for (int i = 0; i < size; i++){char cur = input[i];if (cur == '+' || cur == '*' || cur == '-'){vector<int> result1 = diffWaysToCompute(input.substr(0, i));vector<int> result2 = diffWaysToCompute(input.substr(i + 1));for (auto n1 : result1){for (auto n2 : result2){if (cur == '+') res.emplace_back(n1 + n2);else if (cur == '-') res.emplace_back(n1 - n2);else res.emplace_back(n1 * n2);}}}}if (res.empty()) res.emplace_back(stoi(input));return res;}};