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241. Different Ways to Add Parentheses （算法第二周）

来源：年旅网

Description：

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 for adding this problem and creating all test cases.

to see which companies asked this question.

算法课的第二周学习的是分治法，也就是化繁为简，这道题起初看是没有一点思路的，这是怎么加这么多个括号的呀，后来了解到思路，首先对每一个运算符号进行分离左右两部分，若其中一部分是表达式，则继续遍历分离直至左右两边都是一个数字/符号，接着则通过运算符运算叠加。目前也没有想到更优化的方法。

class Solution {
public:
	vector<int> diffWaysToCompute(string input) {
		vector<int>res;
		int size = input.size();
		for(int i = 0; i < size; i++)
		{
			char c = input[i];
			if(c == '+' || c == '-' || c== '*')
			{
				vector<int> res1 = diffWaysToCompute(input.substr(0,i));
				vector<int> res2 = diffWaysToCompute(input.substr(i+1));
				for(vector<int>::iterator it1 = res1.begin(); it1 != res1.end(); it1++)
					for(vector<int>::iterator it2 = res2.begin(); it2 != res2.end(); it2++)
					{
						if(c == '+')
							res.push_back(*it1 + *it2);
						else if(c == '-')
							res.push_back(*it1 - *it2);
						else if(c == '*')
							res.push_back(*it1 * *it2);	
					}	
			}
		}
		if(res.empty())
		{
			res.push_back(atoi(input.c_str()));
		}
		return res;
	}
};

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