leetcode + 递归，加括号，判断表达式最终结构有哪些

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        char c= '
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class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        char c= '\0';
        for(int i=0; i< input.size(); i++){
            c = input[i];
            if(c=='+' || c=='-' || c=='*'){
                vector<int> result1 = diffWaysToCompute(input.substr(0, i)); //从index0开始，选取i个字符,到index i-1；
                vector<int> result2 = diffWaysToCompute(input.substr(i+1)); //从index i+1开始截取
                for(int r1=0; r1<result1.size(); r1++){
                    for(int r2=0; r2<result2.size(); r2++){
                        if(c=='+'){
                            result.push_back(result1[r1]+ result2[r2]);
                        }
                        else if(c=='-'){
                            result.push_back(result1[r1]- result2[r2]);
                        }
                        else if(c=='*'){
                            result.push_back(result1[r1]*result2[r2]);
                        }
                    }
                }
            }
        }
        if(result.empty()){//也可以是递归终止点
            result.push_back(atoi(input.c_str()));
        }
        return result;
    }
};
 
 
 
  
  

';
        for(int i=0; i< input.size(); i++){
            c = input[i];
            if(c=='+' || c=='-' || c=='*'){
                vector<int> result1 = diffWaysToCompute(input.substr(0, i)); //从index0开始，选取i个字符,到index i-1；
                vector<int> result2 = diffWaysToCompute(input.substr(i+1)); //从index i+1开始截取
                for(int r1=0; r1<result1.size(); r1++){
                    for(int r2=0; r2<result2.size(); r2++){
                        if(c=='+'){
                            result.push_back(result1[r1]+ result2[r2]);
                        }
                        else if(c=='-'){
                            result.push_back(result1[r1]- result2[r2]);
                        }
                        else if(c=='*'){
                            result.push_back(result1[r1]*result2[r2]);
                        }
                    }
                }
            }
        }
        if(result.empty()){//也可以是递归终止点
            result.push_back(atoi(input.c_str()));
        }
        return result;
    }
};
class Solution {
public:
    vector<int>