查找列表中小于给定值的最大整数和（代码跟踪）

Question

Had this problem in a review guide given by CS professor. I ended up solving the problem relatively naively by computing all possible sums of all integers in the list via recursive backtracking in a helper, storing these in a new list, then iterating through that list to find the max sum less than the given number.

We were given the instructor's far more eloquent solution, but I'm having a lot of trouble dissecting it. Hopefully this is an acceptable place to ask for help breaking down what's going on (if not, apologies).

private static int largestSum(List<Integer> list, int n) {
    if (list.isEmpty()) {
        return 0;
    } else {
        int choice = list.get(0);
        list.remove(0);
        // Don't include choice in final sum
        int largest = sumHelper(list, n);
        if (choice < n) {
            // Include choice in final sum
            int included = choice + sumHelper(list, n - choice);
            // Use whichever result is bigger
            largest = Math.max(largest, included);
        }
        // Return list to original state
        list.add(0, choice);
        return largest;
        }
}

I understand the base case (we're iterating through the list by popping off the first element and recursing on the remainder, so when we hit the base case, we stop recursing and don't want to add anything hence the zero; it also addresses empty list). However, I'm still super unclear about the recursive case. I've stared at the code and tried to trace it for a while now, but wrapping my head around the recursion is proving very difficult. Can someone walk me through it?

Answer 1

Each number in the list can have two possibility:

1. To not be the part of the largest sum. Hence we are removing the number from list and Iterating over remaining list.Since Number is Not selected, NO CHANGES IN THE GIVEN VALUE.

   int largest = sumHelper(list, n);

2. To be the part of the largest sum. Hence we are checking if the number is less than the given value ( only then we will include that number), and including it in largest sum. Then we are Iterating over remaining list, with GIVEN VALUE modified to Value = Value - NumberSelected (Since we are including the current number, so SUM is modified).

   if (choice < n) { int included = choice + sumHelper(list, n - choice); largest = Math.max(largest, included); }