Nested List Weight Sum II

题目

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:
Given the list[[1,1],2,[1,1]], return 8. (four 1's at depth 1, one 2 at depth 2)

Example 2:
Given the list[1,[4,[6]]], return17. (one 1 at depth 3, one 4 at depth 2, and one 6 at depth 1; 1*3 + 4*2 + 6*1 = 17)

解析

利用每一层将当前整数加起来，然后往后遍历多一层就将前面已经加过的数再加一遍

这一题其实挺tricky的，如果说第一道题的关键是记录层次，那么这一题的关键是把这一层的integer sum传到下一层去

//关键点在于把上一层的integer sum传到下一层去，这样的话，接下来还有几层，每一层都会加上这个integer sum,也就等于乘以了它的层数

思路

Method 1:

1. very straightforward, calculate the max depth firstly, then get the weight sum

图解

代码

class Solution {
    int result = 0;
    public int depthSumInverse(List<NestedInteger> nestedList) {
        if (nestedList == null)return 0;
        int depth = maxDepth(nestedList);
        dfs(nestedList, depth);
        return result;
    }

    public int maxDepth(List<NestedInteger> nestedList) {
        int depth = 0;
        for (NestedInteger n : nestedList) {
            if (!n.isInteger()) {
                depth = Math.max(maxDepth(n.getList()), depth);
            }
        }
        return depth + 1;
    }

    public  dfs(List<NestedInteger> nestedList, int depth) {
        for (NestedInteger level : nestedList) {
            if (level.isInteger()) {
                result += level.getInteger() * depth;
            } else {
                dfs(level.getList(), depth - 1);
            }
        }
    }
}

Method 2:

1. The idea is to pass the current found integer sum into the next level of recursion, and return it back again. Then we don't have to count max depth