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]

discussion中有个解法非常清楚，就是递归，分治的思想。

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 == '*') {
vector<int> resultL = diffWaysToCompute(input.substr(0, i));
vector<int> resultR = diffWaysToCompute(input.substr(i+1));
for (auto j : resultL) {
for (auto k : resultR) {
if (cur == '+')
result.push_back(j + k);
else if (cur == '-')
result.push_back(j - k);
else
result.push_back(j * k);
}
}
}
}
if (result.empty()) {
result.push_back(atoi(input.c_str()));
}
return result;
}
};

暂时我还没用DP思想。

这个仍然是递归的套路：拆分成两部分，确定好最分支的情况处理。