leetcode 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"
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) = 10

长见识了，大概知道分治的用法了， 额，求二叉树的高度的思维和 分治很像啊， 快排应该就是分治算法的典型应用

遍历输入字符串，遇到一个运算符就分别遍历运算符的左边和右边，通过这样的递归就能求到该运算符左边和右边的所有可能的值， 递归的终止情况是，字符串中只有数字；

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> ans;
        for(int i=0; i<input.size(); i++){
            if(input[i]=='-' || input[i]=='+' || input[i]=='*'){
                vector<int> l=diffWaysToCompute(input.substr(0, i));
                vector<int> r=diffWaysToCompute(input.substr(i+1));
                for(int j=0; j<l.size(); j++){
                    for(int k=0; k<r.size(); k++){
                        if(input[i]=='-') ans.push_back(l[j]-r[k]);
                        else if(input[i]=='+') ans.push_back(l[j]+r[k]);
                        else if(input[i]=='*') ans.push_back(l[j]*r[k]);
                    }
                }
            }
        }
        if(ans.empty()) ans.push_back(stoi(input));
        return ans;
    }
};