題目鏈接: https://leetcode.com/problems/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]], return 10. (four 1‘s at depth 2, one 2 at depth 1)

Example 2:
Given the list [1,[4,[6]]], return 27. (one 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4*2 + 6*3 = 27)

Time Complexity: O(n). n 是指全部葉子的數目加上dfs走過層數的總數. [[[[[5]]]],[[3]], 1], 3個葉子, dfs一共走了6層. 所以用了 3 + 6 = 9 的時間.

Space: O(D). D 是recursive call用的stack的最大數目, 即是最深的層數, 上面例子最深走過4層, 這裏D = 4.

/**
 * // 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) {
        return depthSum(nestedList, 1);
    }
    
    public int depthSum(List<NestedInteger> nestedList, int weight) {
        int sum = 0;
        for (NestedInteger each : nestedList) {
            if (each.isInteger()) 
                sum += each.getInteger() * weight;
            else sum += depthSum(each.getList(), weight+1);
        }
        return sum;
    }
}

[leetcode] 339. Nested List Weight Sum