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]

含义：计算出对于一串字符串所有加括号下结果的可能

    public List<Integer> diffWaysToCompute(String input) {
//        分而治之。对于输入字符串，若其中有运算符，则将其分为两部分，分别递归计算其值，然后将左值集合与右值集合进行交叉运算，将运算结果放入结果集中；
//        若没有运算符，则直接将字符串转化为整型数放入结果集中。
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < input.length(); i++) {
            Character ch = input.charAt(i);
            if (ch.equals('+') || ch.equals('-') || ch.equals('*')) {
                List<Integer> leftResult = diffWaysToCompute(input.substring(0, i));
                List<Integer> rightResult = diffWaysToCompute(input.substring(i + 1));
                for (Integer left : leftResult)
                    for (Integer right : rightResult) {
                        switch (ch) {
                            case '+':
                                result.add(left + right);
                                break;
                            case '-':
                                result.add(left - right);
                                break;
                            case '*':
                                result.add(left * right);
                                break;
                        }
                    }
            }
        }
        if (result.size() == 0) result.add(Integer.valueOf(input));
        return result;    
    }