241 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".
((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]

详见：https://leetcode.com/problems/different-ways-to-add-parentheses/description/

Java实现：

采用分治算法，分治算法的基本思想是将一个规模为N的问题分解为K个规模较小的子问题，这些子问题相互独立且与原问题性质相同，求出子问题的解，就可得到原问题的解。那么针对本题，以操作符为分界，将字符串分解为较小的两个子字符串，然后依次对两个子字符串进行同样的划分，直到字符串中只含有数字。再根据操作符对两端的数字进行相应的运算。

class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> res = new ArrayList<>();
        for(int i=0;i<input.length();++i){
            char c=input.charAt(i);
            if(c=='+'||c=='-'||c=='*'){
                List<Integer> left=diffWaysToCompute(input.substring(0,i));
                List<Integer> right=diffWaysToCompute(input.substring(i+1));
                for(int j=0;j<left.size();++j){
                    for(int k=0;k<right.size();++k){
                        if(c=='+'){
                            res.add(left.get(j)+right.get(k));
                        }else if(c=='-'){
                            res.add(left.get(j)-right.get(k));
                        }else{
                            res.add(left.get(j)*right.get(k));
                        }
                    }
                }
            }
        }
        if(res.isEmpty()){
            res.add(Integer.parseInt(input));
        }
        return res;
    }
}

C++实现：

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(stoi(input));
        }
        return res;
    }
};

 参考：https://www.cnblogs.com/grandyang/p/4682458.html