241-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]

Credits:

Special thanks to @mithmatt for adding this problem and creating all test cases.

因为需要用到 vector，所以选用了c++，代码如下：

#include <iostream>#include <vector>#include <string>#include <stdlib.h>using namespace std;class Solution {public:        vector<int> diffWaysToCompute(string input) {                vector<int> result;                for(int i = 0; i < input.length(); i++) {                        if(input[i] == '+' || input[i] == '-' || input[i] == '*') {                                vector<int> left = diffWaysToCompute(input.substr(0,i));                                vector<int> right = diffWaysToCompute(input.substr(i+1));                                for(int j = 0 ; j < left.size(); j++) {                                        for(int k = 0; k < right.size(); k++) {                                                if(input[i] == '+') {                                                        result.push_back(left[j] + right[k]);                                                } else if(input[i] == '-') {                                                        result.push_back(left[j] - right[k]);                                                } else {                                                        result.push_back(left[j] * right[k]);                                                }                                        }                                }                        }                }                if(result.empty())                        result.push_back(atoi(input.c_str()));                return result;        }};int main() {        string input;        cin >> input;        Solution s;        vector<int> r = s.diffWaysToCompute(input);        for(vector<int>::iterator iter = r.begin(); iter != r.end(); ++iter) {                cout << *iter << endl;        }        return 0;}

暂时想到这，过第二遍的时候有好的算法再改进。