滴滴的算式翻转题

 //main函数，处理输入，用空格把输入拆分成字符串数字，调用解题函数并输出结果
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int count = in.nextInt();
        in.nextLine();
        String[] factors = in.nextLine().trim().split(" ");
        solve(factors);
        if (factors.length > 0) {
            System.out.print(factors[0]);
            for (int i = 1; i < factors.length; i++) {
                System.out.print(" " + factors[i]);
            }
        }
        System.out.println(" ");
    }

    //解题函数，先处理乘号，再处理加号，时间复杂度O(2n) = O(n)
    //可以优化为只遍历一趟
    private static void solve(String[] factors) {
        solveOperator(factors, "*");
        solveOperator(factors, "+");
    }

    //对单个操作符进行处理
    private static void solveOperator(String[] factors, String operator) {
        int index = 0;
        while (index < factors.length) {
            //找到第一个对应的操作符
            while (index < factors.length && !factors[index].equals(operator)) {
                index++;
            }
            //用来记录和排序操作数
            List<Integer> operands = new ArrayList<>();
            //向operands列表添加连续的operator所连接的操作数
            while (index < factors.length && factors[index].equals(operator)) {
                operands.add(Integer.parseInt(factors[index - 1]));
                index += 2;
            }
            
            //index用于遍历，pointer用于对排序后的结果进行回填
            int pointer;
            // + 的受限条件下，少排序一个数字
            if (operator.equals("+") && index < factors.length && !factors[index].equals("-")) {
                pointer = index - 1;
            } else {
                operands.add(Integer.parseInt(factors[index - 1]));
                pointer = index + 1;
            }
            //排序
            Collections.sort(operands);
            //将排序后的操作数进行回填
            for (int i = 0; i < operands.size(); i++) {
                factors[pointer - (operands.size() * 2 - i * 2)] = operands.get(i).toString();
            }
        }
    }