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]

这道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。这道题跟之前的那道Unique Binary Search Trees II 独一无二的二叉搜索树之二用的方法一样，用递归来解，划分左右子树，递归构造。

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;
    }
};

类似题目：

Remove Invalid Parentheses 移除非法括号

Longest Valid Parentheses

Generate Parentheses

Valid Parentheses

参考资料：

http://www.cnblogs.com/ganganloveu/p/4681439.html

https://leetcode.com/discuss/48488/c-4ms-recursive-method