[leetcode] 241. 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]

解法一：

分为左右subpart，然后用基于recursion的思路去做。这里的base case，就是当剩下一个数字的时候，res中为空，这时候将数字的值push_back到res中。

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> res;
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]=='-') res.push_back(left[j]-right[k]);
else if(input[i]=='+') res.push_back(left[j]+right[k]);
else if(input[i]=='*') res.push_back(left[j]*right[k]);
}
}
}
if(res.empty()) res.push_back(atoi(input.c_str()));
return res;

}
};