题目描述

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]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

分析

使用递归，对于每个运算符号+ – *，将string分成两部分（不包括这个运算符号）。分别计算出两部分的结果，再运用这个运算符号进行运算。

代码

    public List<Integer> diffWaysToCompute(String input) {

        List<Integer> rt = new LinkedList<Integer>();
        int len = input.length();

        for (int i = 0; i < len; i++) {

            if (input.charAt(i) == '-' || input.charAt(i) == '*'
                    || input.charAt(i) == '+') {

                String part1 = input.substring(0, i);
                String part2 = input.substring(i + 1);

                List<Integer> part1Ret = diffWaysToCompute(part1);
                List<Integer> part2Ret = diffWaysToCompute(part2);

                for (Integer p1 : part1Ret) {
                    for (Integer p2 : part2Ret) {
                        int c = 0;
                        switch (input.charAt(i)) {
                        case '+':
                            c = p1 + p2;
                            break;
                        case '-':
                            c = p1 - p2;
                            break;
                        case '*':
                            c = p1 * p2;
                        }
                        rt.add(c);
                    }
                }
            }
        }

        if (rt.size() == 0) {
            rt.add(Integer.valueOf(input));
        }

        return rt;
    }