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]
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; } };
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; } };
Different Ways to Add Parentheses
原文：http://www.cnblogs.com/dylqt/p/4988444.html