241. Different Ways to Add Parentheses-优快云博客

本文介绍了一种算法，用于给定包含数字和运算符的字符串时，计算并返回所有可能的结果。通过递归地考虑每个运算符作为最后一个操作的情况，并使用深度优先搜索来解决这个问题。

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]

解题思路：本题的核心思想是每个运算符都可能是最后运算的操作符。

class Solution {
public:
    int sum=0;
    int compute(int a,int b,char op){
        if(op=='+')return a+b;
        else if(op=='-')return a-b;
        else return a*b;
    }
    vector<int> diffWaysToCompute(string input) {
        unordered_map<string,vector<int> >m;
        return computeWithDp(input,m);
    }
    vector<int> computeWithDp(string input,unordered_map<string,vector<int> >&m){
        vector<int>res;
        int size=input.size();
        for(int i=0;i<size;i++){
            if(input[i]=='+'||input[i]=='-'||input[i]=='*'){
                string str=input.substr(0,i);
                vector<int> l,r;
                if(m.count(str))return m[str];
                else l=diffWaysToCompute(str);
                str=input.substr(i+1);
                if(m.count(str))return m[str];
                else r=diffWaysToCompute(str);
                for(auto x: l)
                    for(auto y:r)
                        res.push_back(compute(x,y,input[i]));
            }
        }
        if(res.empty())
            res.push_back(stoi(input));
        return res;
    }
};