241. 为运算表达式设计优先级

给定一个含有数字和运算符的字符串，为表达式添加括号，改变其运算优先级以求出不同的结果。你需要给出所有可能的组合的结果。有效的运算符号包含 +, - 以及 * 。

示例 1:

输入: "2-1-1"
输出: [0, 2]
解释:
((2-1)-1) = 0
(2-(1-1)) = 2
示例 2:

输入: "2*3-4*5"
输出: [-34, -14, -10, -10, 10]
解释:
(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

来源：力扣（LeetCode）

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> res;
        for (int i = 0; i < input.size(); i++) {
            char ch = input[i];
            if (ch == '+' || ch == '-' || ch == '*') {
                vector<int> a = diffWaysToCompute(input.substr(0, i));
                vector<int> b = diffWaysToCompute(input.substr(i+1 , input.size()));
                for (auto i: a) {
                    for (auto j: b) {
                        if (ch == '+')
                            res.push_back(i + j);
                        else if (ch == '-')
                            res.push_back(i - j);
                        else
                            res.push_back(i * j);
                    }
                }
            }
        }
        if (res.empty())
            res.push_back(atoi(input.c_str()));
　　　　　//c_str:Returns a pointer to an array that contains a null-terminated sequence of characters (i.e., a C-string) representing the current value of the string object.
　　　　　//atoi:Convert string to integer

return res;
    }
};

原文地址：https://www.cnblogs.com/xxxsans/p/13828870.html