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: "23-45"

(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]

一刷
題解：recursion
運算符兩邊的子字符串則為該問題的子問題。然后把子問題的解一一做運算。

public class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> res = new LinkedList<>();
        for(int i=0; i<input.length(); i++){
            char ch = input.charAt(i);
            if(ch == '-' || ch == '*' || ch == '+'){
                String part1 = input.substring(0, i);
                String part2 = input.substring(i+1);
                List<Integer> part1Res = diffWaysToCompute(part1);
                List<Integer> part2Res = diffWaysToCompute(part2);
                for(int p1 : part1Res){
                    for(int p2 : part2Res){
                        int c = 0;
                        switch(ch){
                            case '+': c = p1 + p2;
                                break;
                            case '-': c = p1 - p2;
                                break;
                            case '*': c = p1 * p2;
                                break;
                        }
                        res.add(c);
                    }
                }
            }
        }
        if(res.size() == 0){
            res.add(Integer.valueOf(input));
        }
        return res;
    }
}