Leetcode——241. Different Ways to Add Parentheses

description

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]

solution

这一题挺难的，反正思想就是递归求解。递归可以把前后两部分从i位置处分开，然后依次计算前一部分和后一部分的各种可能的解。
递归的设计不太好想！

class Solution {public:    vector<int> diffWaysToCompute(string input) {         if(input.size()==0) return {};         vector<int>res;         for(int i=0;i<input.size();i++)         {             if(input[i]!='*'&&input[i]!='+'&&input[i]!='-') continue;             vector<int> tem1=diffWaysToCompute(input.substr(0,i));             auto tem2=diffWaysToCompute(input.substr(i+1));             for(auto vec1:tem1)                for(auto vec2:tem2)                {                    switch(input[i])                    {                        case '+':                            res.push_back(vec1+vec2);                            break;                        case '-':                            res.push_back(vec1-vec2);                            break;                        case '*':                            res.push_back(vec1*vec2);                            break;                        default:                            break;                    }                }         }         if(res.size()==0)            return vector<int>{stoi(input)};        else            return res;    }};

自己写的不对的代码：
每次求取两个数的运算，把n个数的运算转化为n-1个，依次递减到2个数的运算。
但是存在重复的问题：
写了半天，也粘贴出来：

class Num241 {//递归求解public:    vector<int> diffWaysToCompute(string input) {        vector<int> res;        vector<int> num;        vector<char>ope;        int i = 0;        while (i<input.length())        {            int a = 0;            int sign = 1;            while (i<input.length() && (isdigit(input[i]) || (input[i] == '-' && (!isdigit(input[i + 1])))))//类似2-1这种            {                if (input[i] == '-') sign = -1;                else                    a = a * 10 + (input[i] - '0');                i++;            }            num.push_back(a);            if (i<input.length())                ope.push_back(input[i]);            i++;        }        int count = num.size();        helper(res, num, ope);        return res;        //unordered_map<int,int> resres;        //unordered_map<int, int>::iterator it;        //for (int i = 0;i < res.size();i++)        //{        //      resres[res[i]] = 1;        //}        //vector<int> newres;        //for (it = resres.begin();it != resres.end();it++)        //  newres.push_back(it->first);        //return newres;    }private:    void helper(vector<int>& res, vector<int> num, vector<char>ope)    {        if (num.size() == 2)        {            int a = num[0], b = num[1];            cout <<"res"<< a << " " << b <<" "<< ope[0]<<endl;            switch (ope[0])            {            case '+':                res.push_back(a + b);                break;            case '-':                res.push_back(a - b);                break;            case '*':                res.push_back(a*b);                break;            default:                break;            }            //return;        }        else        {            for (int i = 0; i<num.size() - 1; i++)//            {                int temp = 0;                int a = num[i], b = num[i + 1];                cout << a << " " << b << endl;                switch (ope[i])                {                case '+':                    temp = a + b;                    break;                case '-':                    temp = a - b;                    break;                case '*':                    temp = a*b;                    break;                default:                    break;                }                vector<int> num_copy(num);                vector<char> ope_copy(ope);                num_copy[i] = temp;                num_copy.erase(num_copy.begin() + i + 1);                ope_copy.erase(ope_copy.begin() + i);                helper(res, num_copy, ope_copy);            }        }    }};

另外一种递归方法（去重复）：

class Solution {public:    vector<int> diffWaysToCompute(string input) {        unordered_map<string, vector<int>> dpMap;        return computeWithDP(input, dpMap);    }    vector<int> computeWithDP(string input, unordered_map<string, vector<int>> &dpMap) {        vector<int> result;        int size = input.size();        for (int i = 0; i < size; i++) {            char cur = input[i];            if (cur == '+' || cur == '-' || cur == '*') {                // Split input string into two parts and solve them recursively                vector<int> result1, result2;                string substr = input.substr(0, i);                // check if dpMap has the result for substr                if (dpMap.find(substr) != dpMap.end())                    result1 = dpMap[substr];//没有重复的                else                    result1 = computeWithDP(substr, dpMap);//有重复的，继续递归下去                substr = input.substr(i + 1);                if (dpMap.find(substr) != dpMap.end())                    result2 = dpMap[substr];                else                    result2 = computeWithDP(substr, dpMap);                for (auto n1 : result1) {                    for (auto n2 : result2) {                        if (cur == '+')                            result.push_back(n1 + n2);                        else if (cur == '-')                            result.push_back(n1 - n2);                        else                            result.push_back(n1 * n2);                    }                }            }        }        // if the input string contains only number        if (result.empty())            result.push_back(atoi(input.c_str()));        // save to dpMap        dpMap[input] = result;        return result;    }};