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]

给一个字符串表达式，可以给它任意加括号，求所有表达式的值。
重复元素不用管，因此可以直接搜索。

不含有符号的字符串表示一个数，直接返回只有这个数的容器。

否则根据符号两边的结果相互组合。

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> res;
for(int i = 0; i < input.size(); i++)
{
if(input[i] == '+')
{
for (int a : diffWaysToCompute(input.substr(0, i)))
for (int b : diffWaysToCompute(input.substr(i+1)))
res.push_back(a + b);
}
else if(input[i] == '-')
{
for (int a : diffWaysToCompute(input.substr(0, i)))
for (int b : diffWaysToCompute(input.substr(i+1)))
res.push_back(a - b);
}
else if(input[i] == '*')
{
for (int a : diffWaysToCompute(input.substr(0, i)))
for (int b : diffWaysToCompute(input.substr(i+1)))
res.push_back(a * b);
}
}
if(res.empty()) res.push_back(stoi(input));
return res;
}
};