
1 azh7138m 2018-02-16 12:58:33 +08:00 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 2018-02-16 17:23:04 +08:00 via Android 打开要登陆,登陆完 404 |
5 hotea 2018-02-16 17:47:53 +08:00 |
6 Cbdy 2018-02-16 17:53:26 +08:00 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 2018-02-16 18:08:28 +08:00 via Android 502 |
9 naiba 2018-02-16 18:29:32 +08:00 via Android 你这个叫 LCC 吧 |
10 cheesea 2018-02-16 19:35:29 +08:00 账户不互通,没有讨论。 综上 lcc 打入死牢 |
11 banxi1988 OP |
12 hazhaz 2018-02-16 22:57:16 +08:00 @cheesea 刚开放测试啊,讨论应该是会有的吧,总之不用翻墙了。新年想刷个 666,结果有 665,有 667,结果就是没有 666 |
13 ToT 2018-02-17 04:58:25 +08:00 @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 repesents is: 3 \ 1 The path sum is (3 + 1) = 4. |