算法生成添加到目标的所有数学表达式组合（Java作业/面试）

Question

我试图解决下面的编码挑战问题，但无法在1小时内完成。我对算法的工作原理有一个想法，但我不太确定如何最好地实现它。下面是我的代码和问题

The first 12 digits of pi are 314159265358. We can make these digits into an expression evaluating to 27182 (first 5 digits of e) as follows:

3141 * 5 / 9 * 26 / 5 * 3 - 5 * 8 = 27182

or

3 + 1 - 415 * 92 + 65358 = 27182

Notice that the order of the input digits is not changed. Operators (+,-,/, or *) are simply inserted to create the expression.

Write a function to take a list of numbers and a target, and return all the ways that those numbers can be formed into expressions evaluating to the target

For example:
f("314159265358", 27182) should print:

3 + 1 - 415 * 92 + 65358 = 27182
3 * 1 + 4 * 159 + 26535 + 8 = 27182
3 / 1 + 4 * 159 + 26535 + 8 = 27182
3 * 14 * 15 + 9 + 26535 + 8 = 27182
3141 * 5 / 9 * 26 / 5 * 3 - 5 * 8 = 27182

这个问题很难，因为你可以把数字组合在一起，而你一次也不考虑一个数字。我不确定该如何进行该步骤的组合和递归。请注意，解决方案中没有提供括号，但保留了操作顺序

我的目标是从说开始

{"3"}
then
{"31", "3+1", "3-1", "3*1" "3/1"}
then
{"314", "31+4", "3+1+4", "3-1-4", "31/4", "31*4", "31-4"} etc.

然后每次查看列表中的每个值，看看它是否为目标值。如果是，则将该字符串添加到结果列表中

这是我的密码

public static List<String> combinations(String nums, int target)
    {

        List<String> tempResultList = new ArrayList<String>();
        List<String> realResultList = new ArrayList<String>();
        String originalNum = Character.toString(nums.charAt(0));


        for (int i = 0; i < nums.length(); i++)
        {
            if (i > 0)
            {
                originalNum += nums.charAt(i); //start off with a new number to decompose
            }
            tempResultList.add(originalNum);
            char[] originalNumCharArray = originalNum.toCharArray();
            for (int j = 0; j < originalNumCharArray.length; j++)
            {
                //go through every character to find the combinations?
                // maybe recursion here instead of iterative would be easier...
            }
            for (String s : tempResultList)
            {
                //try to evaluate
                int temp = 0;
               if (s.contains("*") || s.contains("/") || s.contains("+") || s.contains("-"))
               {
                  //evaluate expression
               } else {
                   //just a number
               }
                if (temp == target)
                {
                    realResultList.add(s);
                }

            }
         tempResultList.clear();
        }
        return realResultList;
    }

有人能帮我解决这个问题吗？正在寻找包含编码的答案，因为我需要帮助生成可能性