Given a nested list of integers, return the sum of all integers in the list weighted by their depth. Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Have you met this question in a real interview?

Yes

Example

Given the list[[1,1],2,[1,1]], return10. (four 1's at depth 2, one 2 at depth 1, 4 * 1 * 2 + 1 * 2 * 1 = 10)
Given the list[1,[4,[6]]], return27. (one 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4_2 + 6_3 = 27)

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *
 *     // @return true if this NestedInteger holds a single integer,
 *     // rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds,
 *     // if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // @return the nested list that this NestedInteger holds,
 *     // if it holds a nested list
 *     // Return null if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
public class Solution {
    public int depthSum(List<NestedInteger> nestedList) {
        // Write your code here
        if (nestedList == null || nestedList.size() == 0) {
            return 0;
        }

        return dfs(1, nestedList);

    }

    private int dfs(int depth, List<NestedInteger> nestedList) {

        int sum = 0;

        for (int i = 0; i < nestedList.size(); i++) {
            if (nestedList.get(i).isInteger()) {
                sum += nestedList.get(i).getInteger() * depth;
                continue;
            }

            sum += dfs(depth + 1, nestedList.get(i).getList());
        }

        return sum;
    }
}

1. non-recursion