241. Different Ways to Add Parentheses

题目：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 1Input: 

"2-1-1"

.

((2-1)-1) = 0
(2-(1-1)) = 2

Output: 

[0, 2]

Example 2Input: 

"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]

翻译：

给定一串数字和运算符，从计算所有不同的可能方式返回所有可能的结果，以组合数字和运算符。 有效的运算符是+， - 和*。实施例1输入：“2-1-1”。（（2-1）-1）= 0（2-（1-1））= 2输出：[0，2]实施例2输入：“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输出：[-34，-14，-10,10-10]

代码展示：#include<map>#include<iostream>#include<string>#include<vector>#include<cstdlib>using namespace std;class Solution {vector<int> f(string str,map< string , vector<int> > &dmap) {int len = str.length();vector<int> v;for(int i = 0; i< len;i++) {if(str[i] == '+' || str[i] == '-' || str[i] == '*') {string subStr1 = str.substr(0,i);vector<int> v1,v2;if(dmap.find(subStr1) != dmap.end()) v1 = dmap[subStr1];else v1 = f(subStr1,dmap);string subStr2 = str.substr(i+1);if(dmap.find(subStr2) != dmap.end()) v2 = dmap[subStr2];else v2 = f(subStr2,dmap);for(int j = 0 ; j < v1.size(); j++) {for(int k = 0 ; k < v2.size(); k++) {switch(str[i]) {case '+':v.push_back(v1[j]+v2[k]);break;case '-':v.push_back(v1[j]-v2[k]);break;default:v.push_back(v1[j]*v2[k]);break;}}}}}if(v.empty()) v.push_back(atoi(str.c_str() ));dmap[str] = v;return v;}public:vector<int> diffWaysToCompute(string input) {map<string, vector<int> > dmap;return f(input,dmap);}};int main() {Solution s;vector<string> strs;strs.push_back("2-1-1");strs.push_back("2*3-4*5");for(int i = 0; i< strs.size() ;i++) {vector<int> v = s.diffWaysToCompute(strs[i]);cout << "{ ";for(int j = 0 ; j< v.size(); j++) {cout << v[j] << ",";}cout << " }\n";}}解题状态：