Description

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.

Different from the previous question where weight is increasing from root to leaf, now the weight is defined from bottom up. i.e., the leaf level integers have weight 1, and the root level integers have the largest weight.

Example 1:

Input: [[1,1],2,[1,1]] Output: 8 Explanation: Four 1’s at depth 1, one 2 at depth 2.

Example 2:

Input: [1,[4,[6]]] Output: 17 Explanation: One 1 at depth 3, one 4 at depth 2, and one 6 at depth 1; 13 + 42 + 6*1 = 17.

Solution

給一個巢狀的list，計算它的加權和。和之前不同在於是從底向上統計的。使用遞迴解決，每次計算這一層的和然後傳遞到下一層的計算中，這樣每次都會加和一次，count到listsum中。

We could iterate this nestedInteger recursively. Pass the sum of this level’s integers to next level. It would be counted for another time.

Code

/**
 * // 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 {
    public int depthSumInverse(List<NestedInteger> nestedList) {
        return helper(nestedList, 0);
    }
    
    private int helper(List<NestedInteger> nestedList, int prevSum){
        int intSum = prevSum;
        List<NestedInteger> nextList = new ArrayList<>();
        
        for (NestedInteger n : nestedList){
            if (n.isInteger()){
                intSum += n.getInteger();
            }
            else{
                nextList.addAll(n.getList());
            }
        }
        
        int listSum = nextList.isEmpty() ? 0 : helper(nextList, intSum);
        return intSum + listSum;
    }
}

Time Complexity: unknow Space Complexity: unknow

Review

addAll() is different from add(), add() just add an item to list, even If you pass a list to it. But addAll() could add a list into it.