339. Nested List Weight Sum

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.

Example 1:
Given the list[[1,1],2,[1,1]], return10. (four 1's at depth 2, one 2 at depth 1)

Example 2:
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)

思路与想法：

Typical dfs problem, pass list and depth as parameter

recursive solution:

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *     // Constructor initializes an empty nested list.
 *     public NestedInteger();
 *
 *     // Constructor initializes a single integer.
 *     public NestedInteger(int value);
 *
 *     // @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();
 *
 *     // Set this NestedInteger to hold a single integer.
 *     public void setInteger(int value);
 *
 *     // Set this NestedInteger to hold a nested list and adds a nested integer to it.
 *     public void add(NestedInteger ni);
 *
 *     // @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();
 * }
 */
class Solution {
    int sum = 0;
    public int depthSum(List<NestedInteger> nestedList) {
        if (nestedList == null || nestedList.size() == 0) {
            return 0;
        }
        dfs(nestedList, 1);
        return sum;
    }

    public void dfs(List<NestedInteger> list, int depth) {
        for (int i = 0; i < list.size(); i++) {
            NestedInteger curr = list.get(i);
            if (curr.isInteger()) {
                sum += depth * curr.getInteger();
            } else {
                dfs(curr.getList(), depth + 1);
            }
        }
    }
}

Time Complexity:

The time complexity of my solution is O(n), n is the number of elements or Integer within the list.

Space Complexity:

O(k) where k is the max level of depth.

Iterative Solution:

    public int depthSum(List<NestedInteger> nestedList) {
        int res = 0;
        Stack<Iterator<NestedInteger>> stk = new Stack<>();
        stk.push (nestedList.iterator());

        while (!stk.isEmpty()) {
            Iterator<NestedInteger> itr = stk.peek ();
            while (itr.hasNext()) {
                NestedInteger n = itr.next();
                if (n.isInteger()) res += n.getInteger() * stk.size ();
                else {
                    stk.push (n.getList().iterator());
                    break;
                }
            }
            if (!stk.peek ().hasNext()) stk.pop ();
        }
        return res;
    }

339.Nested List Weight Sum