Leetcode-241: designing priorities for operational expressions

subject

Topic linking

Give you a string of numbers and operators expression , Combine numbers and operators according to different priorities , Calculate and return the results of all possible combinations . You can In any order Return to the answer .

The generated test case meets its corresponding output value in line with 32 Bit integer range , The number of different results does not exceed 104 .

Example 1：

 Input ：expression = "2-1-1"
 Output ：[0,2]
 explain ：
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2：

 Input ：expression = "2*3-4*5"
 Output ：[-34,-14,-10,-10,10]
 explain ：
(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

Problem solving

Method 1 ：dfs（ Divide and conquer ）

class Solution {
    
public:
    vector<int> diffWaysToCompute(string expression) {
    
        return dfs(0,expression.size()-1,expression);
    }
    vector<int> dfs(int l,int r,string& s){
    
        vector<int> res;
        for(int i=l;i<=r;i++){
    
            if(s[i]>='0'&&s[i]<='9') continue;
            vector<int> l1=dfs(l,i-1,s);// Divide and conquer ： Get the result on the left side of the operator 
            vector<int> l2=dfs(i+1,r,s);// Divide and conquer ： Get the result on the right side of the operator 
            for(int a:l1){
    
                for(int b:l2){
    
                    int cur=0;
                    if(s[i]=='+') cur=a+b;
                    else if(s[i]=='-') cur=a-b;
                    else cur=a*b;
                    res.push_back(cur);
                }
            }
        }
        if(res.empty()){
    
            int num=0;
            for(int i=l;i<=r;i++){
    
                num=num*10+s[i]-'0';
            }
            res.push_back(num);
        }
        return res;
    }
};