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]

递归算出符号两边表达式的所有结果，然后执行操作算出所有结果。最后排序，返回

代码：

class Solution {
public:
vector<int> diffWaysToCompute(string input)
{
vector<int>num;
vector<int>left,right;
int n=input.size();
int flg=0;
for(int i=0;i<n;i++)
{
if(input[i]=='+'||input[i]=='-'||input[i]=='*')
{
left=diffWaysToCompute(input.substr(0,i));
right=diffWaysToCompute(input.substr(i+1,n-i));
for(int j=0;j<left.size();j++)
{
for(int k=0;k<right.size();k++)
{
if(input[i]=='+') num.push_back(left[j]+right[k]);
else if(input[i]=='-') num.push_back(left[j]-right[k]);
else if(input[i]=='*') num.push_back(left[j]*right[k]);
}
}
flg=1;
}
}
if(flg==0)
{
num.push_back(atoi(input.c_str()));
}
sort(num.begin(),num.end());
return num;
}
};