From dbcec85e977fce41e5459ef3d108b29abc98883b Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:22:10 +0800 Subject: [PATCH 1/9] =?UTF-8?q?update=200455.=E5=88=86=E5=8F=91=E9=A5=BC?= =?UTF-8?q?=E5=B9=B2=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82=E5=BA=A6?= =?UTF-8?q?=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0455.分发饼干.md | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/problems/0455.分发饼干.md b/problems/0455.分发饼干.md index 3688b3fd..efadb433 100644 --- a/problems/0455.分发饼干.md +++ b/problems/0455.分发饼干.md @@ -56,8 +56,6 @@ C++代码整体如下: ```CPP // 版本一 -// 时间复杂度:O(nlogn) -// 空间复杂度:O(1) class Solution { public: int findContentChildren(vector& g, vector& s) { @@ -75,6 +73,9 @@ public: } }; ``` +* 时间复杂度:O(nlogn) +* 空间复杂度:O(1) + 从代码中可以看出我用了一个 index 来控制饼干数组的遍历,遍历饼干并没有再起一个 for 循环,而是采用自减的方式,这也是常用的技巧。 @@ -118,6 +119,9 @@ public: } }; ``` +* 时间复杂度:O(nlogn) +* 空间复杂度:O(1) + 细心的录友可以发现,这种写法,两个循环的顺序改变了,先遍历的饼干,在遍历的胃口,这是因为遍历顺序变了,我们是从小到大遍历。 From 80006ad61592e0268fd4a8857a8a8b687278c5b3 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:28:09 +0800 Subject: [PATCH 2/9] =?UTF-8?q?update=200053.=E6=9C=80=E5=A4=A7=E5=AD=90?= =?UTF-8?q?=E5=BA=8F=E5=92=8C=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 1de68041..48f8be29 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -24,8 +24,6 @@ 暴力解法的思路,第一层 for 就是设置起始位置,第二层 for 循环遍历数组寻找最大值 -- 时间复杂度:O(n^2) -- 空间复杂度:O(1) ```CPP class Solution { @@ -44,6 +42,9 @@ public: } }; ``` +* 时间复杂度:O(n^2) +* 空间复杂度:O(1) + 以上暴力的解法 C++勉强可以过,其他语言就不确定了。 @@ -98,7 +99,6 @@ public: } }; ``` - - 时间复杂度:O(n) - 空间复杂度:O(1) From 85f54a718d022e6a3cb5883fb3d9aff5745a3362 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:30:07 +0800 Subject: [PATCH 3/9] =?UTF-8?q?update=20=20=200055.=E8=B7=B3=E8=B7=83?= =?UTF-8?q?=E6=B8=B8=E6=88=8F=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0055.跳跃游戏.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index ce9a21ec..c187ef8f 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -75,6 +75,10 @@ public: }; ``` +* 时间复杂度: O(n) +* 空间复杂度: O(1) + + ## 总结 这道题目关键点在于:不用拘泥于每次究竟跳几步,而是看覆盖范围,覆盖范围内一定是可以跳过来的,不用管是怎么跳的。 From 4afa9c2a51f3745b1dc6497fd522a607f3174136 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:32:31 +0800 Subject: [PATCH 4/9] =?UTF-8?q?update=20=20=200045.=E8=B7=B3=E8=B7=83?= =?UTF-8?q?=E6=B8=B8=E6=88=8FII=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D?= =?UTF-8?q?=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0045.跳跃游戏II.md | 9 +++++++++ 1 file changed, 9 insertions(+) diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md index c6e4e9b5..558b32b9 100644 --- a/problems/0045.跳跃游戏II.md +++ b/problems/0045.跳跃游戏II.md @@ -88,6 +88,10 @@ public: }; ``` +* 时间复杂度: O(n) +* 空间复杂度: O(1) + + ## 方法二 依然是贪心,思路和方法一差不多,代码可以简洁一些。 @@ -127,6 +131,11 @@ public: }; ``` +* 时间复杂度: O(n) +* 空间复杂度: O(1) + + + 可以看出版本二的代码相对于版本一简化了不少! **其精髓在于控制移动下标 i 只移动到 nums.size() - 2 的位置**,所以移动下标只要遇到当前覆盖最远距离的下标,直接步数加一,不用考虑别的了。 From 38c74c90d793b9637feceec6350e269d9e460a16 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:34:25 +0800 Subject: [PATCH 5/9] =?UTF-8?q?update=20=20=201005.K=E6=AC=A1=E5=8F=96?= =?UTF-8?q?=E5=8F=8D=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E5=92=8C=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1005.K次取反后最大化的数组和.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index 4a080857..27a575c7 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -85,6 +85,10 @@ public: }; ``` +* 时间复杂度: O(nlogn) +* 空间复杂度: O(1) + + ## 总结 贪心的题目如果简单起来,会让人简单到开始怀疑:本来不就应该这么做么?这也算是算法?我认为这不是贪心? From 25bf0859a1e7b68ddfcb83456e9828244e0ee703 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:36:37 +0800 Subject: [PATCH 6/9] =?UTF-8?q?update=20=20=200135.=E5=88=86=E5=8F=91?= =?UTF-8?q?=E7=B3=96=E6=9E=9C=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0135.分发糖果.md | 5 +++++ 1 file changed, 5 insertions(+) diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md index ef79868a..1ba1563f 100644 --- a/problems/0135.分发糖果.md +++ b/problems/0135.分发糖果.md @@ -121,6 +121,11 @@ public: }; ``` +* 时间复杂度: O(n) +* 空间复杂度: O(n) + + + ## 总结 这在leetcode上是一道困难的题目,其难点就在于贪心的策略,如果在考虑局部的时候想两边兼顾,就会顾此失彼。 From 4c7464fa93ca388699b37935e5b56e68b1f6fed2 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:37:48 +0800 Subject: [PATCH 7/9] =?UTF-8?q?update=20=200860.=E6=9F=A0=E6=AA=AC?= =?UTF-8?q?=E6=B0=B4=E6=89=BE=E9=9B=B6=20=EF=BC=9A=E6=B7=BB=E5=8A=A0?= =?UTF-8?q?=E5=A4=8D=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0860.柠檬水找零.md | 3 +++ 1 file changed, 3 insertions(+) diff --git a/problems/0860.柠檬水找零.md b/problems/0860.柠檬水找零.md index cd52adb0..d96e0879 100644 --- a/problems/0860.柠檬水找零.md +++ b/problems/0860.柠檬水找零.md @@ -115,6 +115,9 @@ public: } }; ``` +* 时间复杂度: O(n) +* 空间复杂度: O(1) + ## 总结 From ff77937816b26b6aa88a41fb6b1684837ca1a31d Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:41:40 +0800 Subject: [PATCH 8/9] =?UTF-8?q?update=20=20=200056.=E5=90=88=E5=B9=B6?= =?UTF-8?q?=E5=8C=BA=E9=97=B4=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0056.合并区间.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index ec659108..72dc69c5 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -73,6 +73,10 @@ public: }; ``` +* 时间复杂度: O(nlogn) +* 空间复杂度: O(logn),排序需要的空间开销 + + ## 其他语言版本 From 528c1d4f9e9678b143706178f76d91c0de6947c4 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Tue, 11 Apr 2023 00:49:38 +0800 Subject: [PATCH 9/9] =?UTF-8?q?update=20=20=200968.=E7=9B=91=E6=8E=A7?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D?= =?UTF-8?q?=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0968.监控二叉树.md | 5 +++++ 1 file changed, 5 insertions(+) diff --git a/problems/0968.监控二叉树.md b/problems/0968.监控二叉树.md index 267e7f6e..fdec56ca 100644 --- a/problems/0968.监控二叉树.md +++ b/problems/0968.监控二叉树.md @@ -303,6 +303,11 @@ public: ``` +* 时间复杂度: O(n),需要遍历二叉树上的每个节点 +* 空间复杂度: O(n) + + + 大家可能会惊讶,居然可以这么简短,**其实就是在版本一的基础上,使用else把一些情况直接覆盖掉了**。 在网上关于这道题解可以搜到很多这种神级别的代码,但都没讲不清楚,如果直接看代码的话,指定越看越晕,**所以建议大家对着版本一的代码一步一步来,版本二中看不中用!**。