1
azh7138m 2018 年 2 月 16 日 via Android
前段时间不是有个 php 的过来吐槽 Perl 加法吗,了解一下
|
4
stzz 2018 年 2 月 16 日 via Android
打开要登陆,登陆完 404
|
5
hotea 2018 年 2 月 16 日
|
6
Cbdy 2018 年 2 月 16 日
10 进制转 26 进制?
以前写过一个 10 进制转 62 进制的小方法用来压缩字符串。。。 |
8
0person 2018 年 2 月 16 日 via Android
502
|
9
naiba 2018 年 2 月 16 日 via Android
你这个叫 LCC 吧
|
10
cheesea 2018 年 2 月 16 日 账户不互通,没有讨论。
综上 lcc 打入死牢 |
11
banxi1988 OP |
13
ToT 2018 年 2 月 17 日
@hazhaz
LeetCode 666 Path Sum IV If the depth of a tree is smaller than 5, then this tree can be represented by a list of three-digits integers. For each integer in this list: The hundreds digit represents the depth D of this node, 1 <= D <= 4. The tens digit represents the position P of this node in the level it belongs to, 1 <= P <= 8. The position is the same as that in a full binary tree. The units digit represents the value V of this node, 0 <= V <= 9. Given a list of ascending three-digits integers representing a binary with the depth smaller than 5. You need to return the sum of all paths from the root towards the leaves. Example 1: Input: [113, 215, 221] Output: 12 Explanation: The tree that the list represents is: 3 / \ 5 1 The path sum is (3 + 5) + (3 + 1) = 12. Example 2: Input: [113, 221] Output: 4 Explanation: The tree that the list represents is: 3 \ 1 The path sum is (3 + 1) = 4. |