241. Different Ways to Add Paren

题目链接
tag:

• Medium；

question：
  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"
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2

Example 2:

Input: "23-45"
Output: [-34, -14, -10, -10, 10]
Explanation:
(2(3-(45))) = -34
((23)-(45)) = -14
((2(3-4))5) = -10
(2((3-4)5)) = -10
(((23)-4)5) = 10

思路：
  本题让给我们一个可能含有加减乘的表达式，让我们在任意位置添加括号，求出所有可能表达式的不同值。用递归来解，划分左右部分，递归构造。代码如下：

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(atoi(input.c_str()));
        return res;
    }
};