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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 *.
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]
Solution:
分治法, 根据运算符将字符串分割成left和right，然后merge左右
1 vector<int> diffWaysToCompute(string input) 2 { 3 vector<int> result; 4 5 for (int i = 0; i < input.size(); i++) 6 { 7 if (input[i] == ‘+‘ || input[i] == ‘-‘ || input[i] == ‘*‘) 8 { 9 vector<int> left = diffWaysToCompute(input.substr(0, i)); 10 vector<int> right = diffWaysToCompute(input.substr(i + 1)); 11 for (int j = 0; j < left.size(); j++) 12 { 13 for (int k = 0; k < right.size(); k++) 14 { 15 if (input[i] == ‘+‘) 16 result.push_back(left[j] + right[k]); 17 else if (input[i] == ‘-‘) 18 result.push_back(left[j] - right[k]); 19 else 20 result.push_back(left[j] * right[k]); 21 } 22 } 23 } 24 } 25 26 // in case input is number only 27 if (result.size() == 0) 28 result.push_back(atoi(input.c_str())); 29 30 return result; 31 }
[leetcode] 241. Different Ways to Add Parentheses
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原文地址：http://www.cnblogs.com/ym65536/p/5401177.html
