241. Different Ways to Add Parentheses（重要）

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]

解题思路：将字符串从任一个运算符处分为两部分，计算第一部分的值，计算第二部分的值，然后将两部分的值根据具体的运算符拼接起来。


class Solution {public:vector<int> diffWaysToCompute(string input) {vector<int> res;for (int i = 0; i < input.size(); i++){if (input[i] == '+' || input[i] == '-' || input[i] == '*'){vector<int> left = diffWaysToCompute(input.substr(0, i));vector<int> right = diffWaysToCompute(input.substr(i + 1));for (int j = 0; j < left.size(); j++){for (int k = 0; k < right.size(); k++){if (input[i] == '+'){res.push_back(left[j] + right[k]);}else if (input[i] == '-'){res.push_back(left[j] - right[k]);}else{res.push_back(left[j] * right[k]);}}}}}if (res.empty())res.push_back(stoi(input));//第一个合格的数字return res;}};