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[//]: #(<p>Given the head node of a linked list <code>head</code> ，The first node returning to the linked list to start the ring。 <em>If the linked linked linked，Then return <code>null</code>。</em></p>)
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[//]: #()
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[//]: #(<p>If there is a node in the linked list，Can be tracked continuously <code>next</code> The pointer arrives again，Then there is a ring in the linked list。 To show the ring in the linked list，Internal use of the evaluation system <code>pos</code> Let's represent the position connected to the linked list to the position of the linked list（<strong>Index 0 start</strong>）。if <code>pos</code> yes <code>-1</code>，There is no ring in this linked list。<strong>Notice：<code>pos</code> Pass the non -as a parameter</strong>，仅仅yes为了标识Linked的实际情况。</p>)
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[//]: #()
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[//]: #(<p><strong>Modification is not allowed </strong>Linked。</p>)
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[//]: #()
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[//]: #(<ul> )
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[//]: #(</ul>)
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[//]: #()
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[//]: #(<p> </p>)
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[//]: #()
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[//]: #(<p><strong>Exemplary example 1：</strong></p>)
[//]: #( <li><code>pos</code> Value <code>-1</code> 或者Linked中的一indivualhave效索引</li> )
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[//]: #(</ul>)
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[//]: #
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[//]: #"# topic："
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[//]: #"<p>Given the head node of a linked list <code>head</code> ，The first node returning to the linked list to start the ring。 <em>If the linked linked linked，Then return <code>null</code>。</em></p>"
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[//]: #
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[//]: #"<p>If there is a node in the linked list，Can be tracked continuously <code>next</code> The pointer arrives again，Then there is a ring in the linked list。 To show the ring in the linked list，Internal use of the evaluation system <code>pos</code> Let's represent the position connected to the linked list to the position of the linked list（<strong>Index 0 start</strong>）。if <code>pos</code> yes <code>-1</code>，There is no ring in this linked list。<strong>Notice：<code>pos</code> Pass the non -as a parameter</strong>，仅仅yes为了标识Linked的实际情况。</p>"
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[//]: #
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[//]: #"<p><strong>Modification is not allowed </strong>Linked。</p>"
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[//]: #
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[//]: #"<ul> "
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[//]: #"</ul>"
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[//]: #
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[//]: #"<p> </p>"
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[//]: #
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[//]: #"<p><strong>Exemplary example 1：</strong></p>"
这类Linkedtopic一般都yes使用双pointer法解决的，For example, looking for distance tail K Node、Find a ring entrance、Find the entrance of the public tail, etc.。
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### Algorithm：
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1. Double pointer met for the first time： Set two pointers fast，slow 指向Linked头部 head，fast Walk per round 2 step，slow Walk per round 1 step；
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a. First result： fast pointer走过Linked末端，说明Linked无环，直接return null；
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a. First result： fast pointer走过Linked末端，说明Linked无环，直接return null；
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.TIPS: If there is a ring，Two pointers will definitely meet。Because every time I go 1 wheel，fast and slow Pitch +1，fast The event will catch up slow；
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b. Second result： whenfast == slowhour， Two pointers in the ring First encounter 。下面分析此hourfast and slowPassing step数关系 ：
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.TIPS: If there is a ring，Two pointers will definitely meet。Because every time I go 1 wheel，fast and slow Pitch +1，fast The event will catch up slow；
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b. Second result： whenfast == slowhour， Two pointers in the ring First encounter 。下面分析此hourfast and slowPassing step数关系 ：
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.设Linked共have a+b Node，in Linked头部到Linked入口 have a Node（不计Linked入口节点）， Linked环 have b Node（这里需要Notice，a and b yes未知数，例如图解上Linked a=4 , b=5）；Set two pointers分别走了 f，s step，则have：
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a. fast 走的step数yesslowstep数的 2 Double，Right now f=2s；（Analyze： fast Walk per round 2 step）
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b.fast Compare slowGo more n Length of a ring，Right now f=s+nb；（ Analyze： Double pointers have gone through a step，然后在环Inside绕圈直到重合，重合hour fast Compare slow Go 环的长度整数Double ）；
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.The above two types are reduced：f=2nb，s=nb，Right nowfastandslow The pointer left separately 2n，n indivual Circumference of the ring （Notice： n yes未知数，不同Linked的情况不同）。
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.设Linked共have a+b Node，in Linked头部到Linked入口 have a Node（不计Linked入口节点）， Linked环 have b Node（这里需要Notice，a and b yes未知数，例如图解上Linked a=4 , b=5）；Set two pointers分别走了 f，s step，则have：
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a. fast 走的step数yesslowstep数的 2 Double，Right now f=2s；（Analyze： fast Walk per round 2 step）
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b.fast Compare slowGo more n Length of a ring，Right now f=s+nb；（ Analyze： Double pointers have gone through a step，然后在环Inside绕圈直到重合，重合hour fast Compare slow Go 环的长度整数Double ）；
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.The above two types are reduced：f=2nb，s=nb，Right nowfastandslow The pointer left separately 2n，n indivual Circumference of the ring （Notice： n yes未知数，不同Linked的情况不同）。
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<!--more-->
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2. Current situation analysis：
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.if让pointer从Linked头部一直向前走并统计step数k，那么所have 走到Linked入口节点hour的step数 yes：k=a+nb（Leave first a step到入口节点，Over time 1 Ring（ b step）I will go to the entrance node again）。
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..Current，slow pointerPassingstep数为 nb step。therefore，We just find a way slow Go again a step停下来，You can get to the entrance to the ring。
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.if让pointer从Linked头部一直向前走并统计step数k，那么所have 走到Linked入口节点hour的step数 yes：k=a+nb（Leave first a step到入口节点，Over time 1 Ring（ b step）I will go to the entrance node again）。
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..Current，slow pointerPassingstep数为 nb step。therefore，We just find a way slow Go again a step停下来，You can get to the entrance to the ring。
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...但yes我们不知道 a Value，what to do？依然yes使用双pointer法。我们构建一indivualpointer，此pointer需要have以下性质：此pointerandslow Go forward together a step后，The two nodes at the entrance re -coincide。So where does it need to get to the entrance node? a step？答案yesLinked头部head。
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3. Double pointer met for the second time：
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.slowpointer Unchanged position ，Willfastpointer重新 指向Linked头部节点 ；slowandfast同hour每wheel向前走 1 step；
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.slowpointer Unchanged position ，Willfastpointer重新 指向Linked头部节点 ；slowandfast同hour每wheel向前走 1 step；
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..TIPS：此hour f=0，s=nb ；
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...when fast pointer走到f=a stephour，slow pointer走到steps=a+nb，此hour 两pointer重合，并同hour指向Linked环入口 。
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..TIPS：此hour f=0，s=nb ；
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...when fast pointer走到f=a stephour，slow pointer走到steps=a+nb，此hour 两pointer重合，并同hour指向Linked环入口 。
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4. returnslowpointer指向的节点。
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### Complexity analysis：
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hour间复杂度 O(N) ：In the second encounter，慢pointer须走step数 a<a+b；First encounter中，慢pointer须走step数 a+b−x<a+b，in x 为双pointer重合点and环入口距离；therefore总体为线性复杂度；
Copy file name to clipboardExpand all lines: app/docs/CommunityShare/Leetcode/6323. 将钱分给最多的儿童_translated.md
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---
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title: 6323. Child that divides money the most.md
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date: '2024.01.01 0:00'
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date: "2024.01.01 0:00"
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tags:
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- - Python
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- - answer
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abbrlink: b9130c0e
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docId: kw44if3s2zi4w2gs1gfhxvoz
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---
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# topic：
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I met for the first time：'2023/3/19-16:51'
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[6323. Child that divides money the most.md](https://leetcode.cn/problems/distribute-money-to-maximum-children/)
@@ -38,8 +40,10 @@ if money>8×children，So children−1 A child obtained 8 Dollar，剩下的一A
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if money=8×children−4，So children−2 A child obtained 8 Dollar，The remaining two children sharing the rest 12 Dollar（As long as not 4, 8 Dollar就行），return children−2。
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if，We assumed that there are x A child obtained 8 Dollar，Then the rest of the money is money−8×x，As long as it is guaranteed to be greater than equal to the remaining number of children children−x，You can satisfy the meaning。therefore，We just need to ask x Maximum value，That is the answer。
Copy file name to clipboardExpand all lines: app/docs/CommunityShare/Leetcode/994.腐烂的橘子_translated.md
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title: 994.Rotten orange.md
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date: 2024.05.14 0:00
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date: "2024.05.14 0:00"
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tags:
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- Python
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- BFS
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- Bilateral queue
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abbrlink: 56e64fdd
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docId: axhoyzdtxoc82q58j1os57c8
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---
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# topic：
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"""
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<p>Given <code>m x n</code> grid
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<metacharset="UTF-8" /> <code>grid</code> middle，Each cell can have one of the following three values：</p>
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# Thought：
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这个问题可以用Priority search（BFS）To solve。We need to track the spread of rotten oranges，Record time，And check if there is a fresh orange that cannot be rotten。The initial idea of the original idea：
![github-actions[bot]](https://avatars.githubusercontent.com/in/15368?v=4&size=40){kind=avatar}
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