241. Different Ways to Add Parentheses

241. Different Ways to Add Parentheses

题目：

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]

代码如下：

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        for (int  i = 0; i < input.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 == '+')
                            result.push_back(n1 + n2);
                        if (cur == '-')
                            result.push_back(n1 - n2);
                        if (cur == '*')
                            result.push_back(n1 * n2);
                    }
                }
            }
            //若等式不包含运算符，即只含一个数字
        }
        if (result.empty()) 
            result.push_back(atoi(input.c_str()));
        return result;
    }
};

解题思路：

• 本题采用递归 思路；
• 将每个表达式从符号‘+’‘-’‘*’处递归的分解为括号内外两部分计算，最终变为含一个运算符的等式即可。