241. Different Ways to Add Parentheses241. 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"
Output: [0, 2]
Explanation: 
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2:

Input: "2*3-4*5" Output: [-34, -14, -10, -10, 10] Explanation:  (2*(3-(4*5))) = -34  ((2*3)-(4*5)) = -14  ((2*(3-4))*5) = -10  (2*((3-4)*5)) = -10  (((2*3)-4)*5) = 10241. Different Ways to Add Parentheses

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给定一个包含+-*的运算字符串，求所有加括号的方案计算到的值。
采用分治法的思路，尝试在每个操作符分成两个部分，这就相当于分别把操作符左右两边括起来单独计算。然后left和right分开进行递归，最后merge。递归的终止条件是当字符串中不包含操作符时，直接返回这个字符串对应的数值。
分治法很经典的例子。
代码如下：

class Solution {
private:
    int operate(int a, int b, char c)
    {
        switch (c) {
        case ‘+’:
            return a + b;
        case ‘-‘:
            return a – b;
        case ‘*’:
            return a * b;
        }
    }
public:
    vector<int> diffWaysToCompute(string input)
    {
        int n = input.size();
        vector<int> ans;
        bool found = false;
        for (int i = 0; i < n; ++i) {
            if (input[i] == ‘+’ || input[i] == ‘-‘ || input[i] == ‘*’) {
                found = true;
                vector<int> lefts = diffWaysToCompute(input.substr(0, i));
                vector<int> rights = diffWaysToCompute(input.substr(i + 1));
                for (int j = 0; j < lefts.size(); ++j) {
                    for (int k = 0; k < rights.size(); ++k) {
                        ans.push_back(operate(lefts[j], rights[k], input[i]));
                    }
                }
            }
        }
        if (!found)
            return { atoi(input.c_str()) };
        return ans;
    }
};

本代码提交AC，用时3MS。