題目鏈接
tag:
- Medium二汛；
question：
??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: "23-45"
Output: [-34, -14, -10, -10, 10]
Explanation:
(2(3-(45))) = -34
((23)-(45)) = -14
((2(3-4))5) = -10
(2((3-4)5)) = -10
(((23)-4)5) = 10
思路：
??本題讓給我們一個可能含有加減乘的表達式匹层，讓我們在任意位置添加括號芒澜，求出所有可能表達式的不同值巴粪。用遞歸來解卒废，劃分左右部分徒河，遞歸構(gòu)造多律。代碼如下：
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
res.push_back(left[j] * right[k]);
}
}
}
}
if (res.empty())
res.push_back(atoi(input.c_str()));
return res;
}
};
