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".

((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]

• 递归：对所有操作符递归

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        int size = input.size();
        for (int i = 0; i < size; i++) {
            char cur = input[i];
            if (cur == '+' || cur == '-' || cur == '*') {    //直到检查到操作符
                // Split input string into two parts and solve them recursively
                vector<int> result1 = diffWaysToCompute(input.substr(0, i));
                vector<int> result2 = diffWaysToCompute(input.substr(i+1));
                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()));
        return result;
    }
};

• 动态规划

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];
                else
                    result1 = computeWithDP(substr, dpMap);

                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;
    }
};

• 把每一步的算式和结果都保存到一个map中，这样每次查到之前已经算过时，就不需要重新算了
• 这种题目如果用C写，估计基础功能写完已经累趴了