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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 *.
Input: "2-1-1".
((2-1)-1) = 0 (2-(1-1)) = 2
Output: [0, 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]
class Solution {public:vector<int> diffWaysToCompute(string input) {vector<int> result;for(int i=0;i<input.size();i++){char ch =input[i];if(ch==‘+‘||ch==‘-‘||ch==‘*‘){//递归对字符串0~i-1求计算的结果vector<int> result1=diffWaysToCompute(input.substr(0,i));//递归对字符串i+1~n-1求计算的结果vector<int> result2=diffWaysToCompute(input.substr(i+1));//对所有的结果，分别配对求与ch对应的计算结果，写入result之中for(auto x:result1){for(auto y:result2){if(ch==‘+‘)result.push_back(x+y);else if(ch==‘-‘)result.push_back(x-y);elseresult.push_back(x*y);}}}}//如果result.size()==0,标明子串中只有数字，将其转换为数字，写入result之中if(result.size()==0){result.push_back(atoi(input.c_str()));}//返回resultreturn result;}};
class Solution {private:unordered_map<string,vector<int>> map;public:vector<int> diffWaysToCompute(string input) {vector<int> result;for(int i=0;i<input.size();i++){char ch =input[i];if(ch==‘+‘||ch==‘-‘||ch==‘*‘){//递归对字符串0~i-1求计算的结果vector<int> result1= map.find(input.substr(0,i))==map.end()? diffWaysToCompute(input.substr(0,i)):map[input.substr(0,i)];//递归对字符串i+1~n-1求计算的结果vector<int> result2= map.find(input.substr(i+1))==map.end()? diffWaysToCompute(input.substr(i+1)) :map[input.substr(i+1)];//对所有的结果，分别配对求与ch对应的计算结果，写入result之中for(auto x:result1){for(auto y:result2){if(ch==‘+‘)result.push_back(x+y);else if(ch==‘-‘)result.push_back(x-y);elseresult.push_back(x*y);}}}}//如果result.size()==0,标明子串中只有数字，将其转换为数字，写入result之中if(result.size()==0){result.push_back(atoi(input.c_str()));}map[input]=result;//返回resultreturn result;}};
241.Different Ways to Add Parentheses
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原文地址：http://www.cnblogs.com/zhoudayang/p/5008038.html
