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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.
与Unique Binary Search Trees II思路类似，可以对照看。
本题参考Gcdofree的做法
左右子串分别计算所有可能，然后全排列。
class Solution { public: vector<int> diffWaysToCompute(string input) { vector<int> ret; for(int i = 0; i < input.size(); i ++) { if(input[i] == ‘+‘ || input[i] == ‘-‘ || input[i] == ‘*‘) { vector<int> left = diffWaysToCompute(input.substr(0, i)); vector<int> right = diffWaysToCompute(input.substr(i+1)); for(int j = 0; j < left.size(); j ++) { for(int k = 0; k < right.size(); k ++) { if(input[i] == ‘+‘) ret.push_back(left[j] + right[k]); else if(input[i] == ‘-‘) ret.push_back(left[j] - right[k]); else ret.push_back(left[j] * right[k]); } } } } if(ret.empty()) ret.push_back(atoi(input.c_str())); return ret; } };
【LeetCode】241. Different Ways to Add Parentheses
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原文地址：http://www.cnblogs.com/ganganloveu/p/4681439.html
