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…

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 [leetcode]339. Nested List Wei…

原题链接在这里:https://leetcode.com/problems/nested-list-weight-sum-ii/description/ 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 a…

［抄题］: 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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcode-cn.com/problems/nested-list-weight-sum-ii/ 题目描述 Given a nested list of integers, return the sum of all integers in the list weighted by their depth.…

这个题做了一个多小时,好傻逼. 显而易见计算的话必须知道当前层是第几层,因为要乘权重,想要知道是第几层又必须知道最高是几层.. 用了好久是因为想ONE PASS,尝试过遍历的时候构建STACK,通过和顶端的距离来判定层数,但是其实最后POP的过程相当于又遍历了一次.而且STACK无法O(1) access,换成LIST需要手动来维持顺序. 最终放弃了,看答案..发现都是先遍历一次得到最高权重......仔细想想似乎1-PASS可以用MAP实现,但是每次发现新高权重,就必须更新以前所有的,貌似没必…

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…

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…

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…

［抄题］: 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]], r…