题目描述

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.

样例

Example 1:

Input: [[1,1],2,[1,1]]
Output: 10 
Explanation: Four 1's at depth 2, one 2 at depth 1.
Example 2:

Input: [1,[4,[6]]]
Output: 27 
Explanation: One 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4*2 + 6*3 = 27.

算法1

(DFS) $O(n)$

top down DFS. 每层累加当前数与depth的乘积

时间复杂度

时间复杂度O（n）
空间复杂度O（n），recursion需要分配stack

参考文献

C++ 代码

class Solution {
public:
    int depthSum(vector<NestedInteger>& nestedList) {
        return helper(nestedList,1);
    }
    int helper(vector<NestedInteger>& list, int depth)
    {
        int res=0;
        for(auto a:list)
        {
            res+=a.isInteger()? a.getInteger()*depth: helper(a.getList(),depth+1);
        }
        return res;
    }
};

算法2

(BFS) $O(n)$

层序遍历，每层累加数值与depth的乘积放入res中，非数值的nestedList放入tmp数组中，进入下一层遍历前update原数组并depth++

时间复杂度

时间复杂度O（n）
空间复杂度O（n），tmp数组

参考文献

C++ 代码

class Solution {
public:
    int depthSum(vector<NestedInteger>& nestedList) {
        int res=0, depth=1;

        while(!nestedList.empty())
        {
            vector<NestedInteger> tmp;
            for(auto node: nestedList)
            {
                if(node.isInteger())
                {
                    res+=node.getInteger()*depth;
                }
                else
                {
                    for(auto list: node.getList())
                        tmp.push_back(list);
                }
            }
            depth++;
            nestedList=tmp;
        }
        return res;
    }
};