标签：
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.
先想到是就是递归了，下面是AC代码。可能有更好的解法，想到了再回来补充。
1 class Solution { 2 public: 3 int compute(int a, int b, char op) { 4 switch (op) { 5 case ‘+‘: return a + b; 6 case ‘-‘: return a - b; 7 case ‘*‘: return a * b; 8 } 9 return 1; 10 } 11 vector<int> diffWaysToCompute(string input) { 12 int val = 0, idx = 0; 13 while (idx < input.length() && isdigit(input[idx])) { 14 val *= 10; 15 val += input[idx++] - ‘0‘; 16 } 17 if (idx == input.length()) return {val}; 18 vector<int> res; 19 vector<int> left, right; 20 for (int i = 0; i < input.length(); ++i) { 21 if (!isdigit(input[i])) { 22 left = diffWaysToCompute(input.substr(0, i)); 23 right = diffWaysToCompute(input.substr(i + 1, input.length() -1 - i)); 24 for (int j = 0; j < left.size(); ++j) { 25 for (int k = 0; k < right.size(); ++k) { 26 res.push_back(compute(left[j], right[k], input[i])); 27 } 28 } 29 } 30 } 31 return res; 32 } 33 };
[LeetCode] Different Ways to Add Parentheses
标签：
原文地址：http://www.cnblogs.com/easonliu/p/4681493.html
