title
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"
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation:
(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
solution
递归 分治
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
int size = input.size();
for (int i = 0; i < size; i++) {
char cur = input[i];
if (cur == '+' || cur == '-' || cur == '*') {
// Split input string into two parts and solve them recursively
vector<int> result1 = diffWaysToCompute(input.substr(0, i));
vector<int> result2 = diffWaysToCompute(input.substr(i+1));
for (auto n1 : result1) {
for (auto n2 : result2) {
if (cur == '+')
result.push_back(n1 + n2);
else if (cur == '-')
result.push_back(n1 - n2);
else
result.push_back(n1 * n2);
}
}
}
}
// if the input string contains only number
if (result.empty())
result.push_back(stoi(input));
return result;
}
};