https://leetcode.com/problems/nested-list-weight-sum/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.

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)

Sol:

 
 
Because the input is nested, it is natural to think about the problem in a recursive way. We go through the list of nested integers one by one, keeping track of the current depth d. If a nested integer is an integer n, we calculate its sum as time n×d. If the nested integer is a list, we calculate the sum of this list recursively using the same process but with depth d+1.
 
 
 

Complexity Analysis

The algorithm takes O(N) time, where N is the total number of nested elements in the input list. For example, the list [ [[[[1]]]], 2 ] contains 4 nested lists and 2 nested integers (1and 2), so N=6.

In terms of space, at most O(D) recursive calls are placed on the stack, where D is the maximum level of nesting in the input. For example, D=2 for the input [[1,1],2,[1,1]], and D=3 for the input [1,[4,[6]]].

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

339. Nested List Weight Sum的更多相关文章

  1. 【leetcode】339. Nested List Weight Sum

    原题 Given a nested list of integers, return the sum of all integers in the list weighted by their dep ...

  2. LeetCode 339. Nested List Weight Sum

    原题链接在这里:https://leetcode.com/problems/nested-list-weight-sum/ 题目: Given a nested list of integers, r ...

  3. LeetCode 339. Nested List Weight Sum (嵌套列表重和)$

    Given a nested list of integers, return the sum of all integers in the list weighted by their depth. ...

  4. [leetcode]339. Nested List Weight Sum嵌套列表加权和

    Given a nested list of integers, return the sum of all integers in the list weighted by their depth. ...

  5. 【LeetCode】339. Nested List Weight Sum 解题报告(C++)

    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 dfs 日期 题目地址:https://leetcod ...

  6. [leetcode]364. Nested List Weight Sum II嵌套列表加权和II

    Given a nested list of integers, return the sum of all integers in the list weighted by their depth. ...

  7. [LeetCode] 364. Nested List Weight Sum II_Medium tag:DFS

    Given a nested list of integers, return the sum of all integers in the list weighted by their depth. ...

  8. [LeetCode] Nested List Weight Sum II 嵌套链表权重和之二

    Given a nested list of integers, return the sum of all integers in the list weighted by their depth. ...

  9. LeetCode Nested List Weight Sum

    原题链接在这里:https://leetcode.com/problems/nested-list-weight-sum/ 题目: Given a nested list of integers, r ...

随机推荐

  1. sql server 数据库变成单用户模式的恢复

    USE master;GODECLARE @SQL VARCHAR(MAX);SET @SQL=''SELECT @SQL=@SQL+'; KILL '+RTRIM(SPID)FROM master. ...

  2. postgresql数据库3种程序(rule,trigger ,FUNCTION )

    1. CREATE [ OR REPLACE ] RULE name AS ON event TO table_name [ WHERE condition ] DO [ ALSO | INSTEAD ...

  3. 搭建Tomcat应用服务器、tomcat虚拟主机及Tomcat多实例部署

    一.环境准备 系统版本:CentOS release 6.6 (Final) x86_64 Tomcat版本:tomcat- JDK版本:jdk-8u25-linux-x64 关闭防火墙 软件包下载地 ...

  4. Android SDK + Appium 环境搭建

    一.JDK 安装 说明:JDK是包含了JAVA的运行环境(JVM+Java系统类库)和JAVA工具,所以必须最先安装. 链接: https://pan.baidu.com/s/1NfNK_K7vukF ...

  5. Android 开发 框架系列 百度语音合成

    官方文档:http://ai.baidu.com/docs#/TTS-Android-SDK/6d5d6899 官方百度语音合成控制台:https://cloud.baidu.com/product/ ...

  6. 先 FROM 后 WHERE 再 GROUP BY 再 SELECT 再 order BY

    shutdown -r -t 5office:515740906family-asus:786512915office-T420i:837829568 family-T420:868 638 325 ...

  7. python学习笔记_week28

    heap import heapq import random heap = [] data = list(range(10000)) random.shuffle(data) # for num i ...

  8. 收藏 —— 教你阅读Python开源项目

    https://zhuanlan.zhihu.com/p/22275595?refer=python-cn

  9. PHP 调试打印输出变量

    var_dump ($row); echo "hello"; echo "\n"; print_r ($arr); php 数组 对象 $arr = json_ ...

  10. echart 单选legend 并排序

    java代码 List<Map<String, Object>> AllList = null; JSONArray jsonArray = JSONArray.fromObj ...