https://blog.youkuaiyun.com/papilon_zh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/papilon_/article/details/153623614https://blog.youkuaiyun.com/papilon_/article/details/153623614papilon_Mon, 20 Oct 2025 11:28:48 +0800https://blog.youkuaiyun.com/papilon_/article/details/153395616https://blog.youkuaiyun.com/papilon_/article/details/153395616papilon_Thu, 16 Oct 2025 10:44:04 +0800https://blog.youkuaiyun.com/papilon_/article/details/153344682https://blog.youkuaiyun.com/papilon_/article/details/153344682papilon_Wed, 15 Oct 2025 20:33:10 +0800https://blog.youkuaiyun.com/papilon_/article/details/153250792https://blog.youkuaiyun.com/papilon_/article/details/153250792papilon_Tue, 14 Oct 2025 15:35:33 +0800https://blog.youkuaiyun.com/papilon_/article/details/153194380https://blog.youkuaiyun.com/papilon_/article/details/153194380papilon_Mon, 13 Oct 2025 15:22:28 +0800https://blog.youkuaiyun.com/papilon_/article/details/153051513https://blog.youkuaiyun.com/papilon_/article/details/153051513papilon_Sat, 11 Oct 2025 17:23:38 +0800https://blog.youkuaiyun.com/papilon_/article/details/139189550https://blog.youkuaiyun.com/papilon_/article/details/139189550papilon_Sat, 25 May 2024 10:32:35 +0800= i - left + 1，说明以i-1字符为结尾的无重复最长子串中包含了i字符的重复字符，又因为tmp表示的是i-1无重复，所以tmp可以无痛（隐含的left指针就是i字符前一次出现的地方+1）更新为i - left + 1；4. 根据map中的记录，找到上一个与i字符相同的位置（这里要注意，因为是顺序遍历i，所以在考虑i时，i之前的字符都已经被考虑过了）。]]>https://blog.youkuaiyun.com/papilon_/article/details/135909860https://blog.youkuaiyun.com/papilon_/article/details/135909860papilon_Tue, 30 Jan 2024 10:18:50 +0800https://blog.youkuaiyun.com/papilon_/article/details/135697794https://blog.youkuaiyun.com/papilon_/article/details/135697794papilon_Fri, 19 Jan 2024 15:40:44 +0800https://blog.youkuaiyun.com/papilon_/article/details/135649336https://blog.youkuaiyun.com/papilon_/article/details/135649336papilon_Wed, 17 Jan 2024 17:03:05 +0800https://blog.youkuaiyun.com/papilon_/article/details/135623177https://blog.youkuaiyun.com/papilon_/article/details/135623177papilon_Tue, 16 Jan 2024 19:36:53 +0800https://blog.youkuaiyun.com/papilon_/article/details/135599577https://blog.youkuaiyun.com/papilon_/article/details/135599577papilon_Mon, 15 Jan 2024 16:25:43 +0800https://blog.youkuaiyun.com/papilon_/article/details/135554660https://blog.youkuaiyun.com/papilon_/article/details/135554660papilon_Fri, 12 Jan 2024 19:59:53 +0800https://blog.youkuaiyun.com/papilon_/article/details/135472478https://blog.youkuaiyun.com/papilon_/article/details/135472478papilon_Tue, 09 Jan 2024 19:14:09 +0800https://blog.youkuaiyun.com/papilon_/article/details/135449302https://blog.youkuaiyun.com/papilon_/article/details/135449302papilon_Mon, 08 Jan 2024 11:35:37 +0800https://blog.youkuaiyun.com/papilon_/article/details/135421663https://blog.youkuaiyun.com/papilon_/article/details/135421663papilon_Sat, 06 Jan 2024 10:02:27 +0800 result，push到vector> results。参数：（[1, 9]中的某个数充当遍历）起点start，还剩几个数k，值还差多少n。普普通通组合问题，用回溯，最麻烦的地方是2~9所对应的字母如何映射，我这里的笨办法是分别考虑。]]>https://blog.youkuaiyun.com/papilon_/article/details/135410753https://blog.youkuaiyun.com/papilon_/article/details/135410753papilon_Fri, 05 Jan 2024 15:48:46 +0800 result 中的数恰好够最初的k个，或者当前递归要找0个数。在[1, n]中找k个数的组合，一般的想法就是取区间首项作为第一个数，然后在剩余数中找(k - 1)个数。然后取区间第二项作第一个数。1. 返回值和参数。参数：区间起始值start，区间结束值n，要找k个数。]]>https://blog.youkuaiyun.com/papilon_/article/details/135378068https://blog.youkuaiyun.com/papilon_/article/details/135378068papilon_Thu, 04 Jan 2024 12:32:12 +0800val < low && root->val >= low)此时，root根本不是要删除的，而删除pre比较困难，而且可能存在root是pre右子树中的值这种情况，处理起来更麻烦。针对情况3，说明要保留的部分树的根节点，就是rootbenshen。针对情况3，要在root树中，假如某个节点恰好小于low，则要删除该节点及其左子树，即令其右子树替代他；]]>https://blog.youkuaiyun.com/papilon_/article/details/135354573https://blog.youkuaiyun.com/papilon_/article/details/135354573papilon_Wed, 03 Jan 2024 11:34:14 +0800https://blog.youkuaiyun.com/papilon_/article/details/135331803https://blog.youkuaiyun.com/papilon_/article/details/135331803papilon_Tue, 02 Jan 2024 16:19:29 +0800