From 4856f9786b82c9ca54f2759679e1ef7c9e7a34b0 Mon Sep 17 00:00:00 2001
From: PeixiZ <96801981+PeixiZ@users.noreply.github.com>
Date: Tue, 10 Jan 2023 15:49:23 +0800
Subject: [PATCH 1/4] =?UTF-8?q?=E6=B7=BB=E5=8A=A0C#=E7=A8=8B=E5=BA=8F?=
=?UTF-8?q?=EF=BC=8C=E4=BD=BF=E7=94=A8=E4=B8=89=E6=8C=87=E9=92=88=E6=84=9F?=
=?UTF-8?q?=E8=A7=89=E4=BC=9A=E6=9B=B4=E5=BD=A2=E8=B1=A1=E8=A1=A8=E8=BF=B0?=
=?UTF-8?q?=E9=93=BE=E8=A1=A8=E7=BF=BB=E8=BD=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0206.翻转链表.md | 37 +++++++++++++++++++++++++++++++++++
1 file changed, 37 insertions(+)
diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md
index 44146bb4..7db80fe1 100644
--- a/problems/0206.翻转链表.md
+++ b/problems/0206.翻转链表.md
@@ -628,6 +628,43 @@ impl Solution {
}
}
```
+C#:
+三指针法， 感觉会更直观：
+
+```cs
+public LinkNumbers Reverse()
+{
+ ///用三指针，写的过程中能够弥补二指针在翻转过程中的想象
+ LinkNumbers pre = null;
+ var move = root;
+ var next = root;
+
+ while (next != null)
+ {
+ next = next.linknext;
+ move.linknext = pre;
+ pre = move;
+ move = next;
+ }
+ root = pre;
+ return root;
+}
+
+///LinkNumbers的定义
+public class LinkNumbers
+{
+ ///
+ /// 链表值
+ ///
+ public int value { get; set; }
+
+ ///
+ /// 链表指针
+ ///
+ public LinkNumbers linknext { get; set; }
+}
+```
+
From 241f3af9a8c74a5a42f53e9f39689fd1ab501655 Mon Sep 17 00:00:00 2001
From: shangcode <61669790+shangcode@users.noreply.github.com>
Date: Wed, 11 Jan 2023 09:25:03 +0800
Subject: [PATCH 2/4] =?UTF-8?q?Update=200034.=E5=9C=A8=E6=8E=92=E5=BA=8F?=
=?UTF-8?q?=E6=95=B0=E7=BB=84=E4=B8=AD=E6=9F=A5=E6=89=BE=E5=85=83=E7=B4=A0?=
=?UTF-8?q?=E7=9A=84=E7=AC=AC=E4=B8=80=E4=B8=AA=E5=92=8C=E6=9C=80=E5=90=8E?=
=?UTF-8?q?=E4=B8=80=E4=B8=AA=E4=BD=8D=E7=BD=AE.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
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去掉了一个 Python 解法中的分号。
---
...数组中查找元素的第一个和最后一个位置.md | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md
index c7ff6dce..7e58a870 100644
--- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md
+++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md
@@ -355,8 +355,8 @@ class Solution:
while left <= right:
middle = left + (right-left) // 2
if nums[middle] >= target: # 寻找左边界，nums[middle] == target的时候更新right
- right = middle - 1;
- leftBoder = right;
+ right = middle - 1
+ leftBoder = right
else:
left = middle + 1
return leftBoder
From 7d822fe65a587f883a5d63c1b72c139239f2cff7 Mon Sep 17 00:00:00 2001
From: El nino <69737612+el-nino2020@users.noreply.github.com>
Date: Sat, 14 Jan 2023 09:18:08 +0800
Subject: [PATCH 3/4] =?UTF-8?q?Update=2020201003=E4=BA=8C=E5=8F=89?=
=?UTF-8?q?=E6=A0=91=E5=91=A8=E6=9C=AB=E6=80=BB=E7=BB=93.md?=
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typo?
---
problems/周总结/20201003二叉树周末总结.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/周总结/20201003二叉树周末总结.md b/problems/周总结/20201003二叉树周末总结.md
index 18bbf37f..5f59b040 100644
--- a/problems/周总结/20201003二叉树周末总结.md
+++ b/problems/周总结/20201003二叉树周末总结.md
@@ -2,7 +2,7 @@
本周赶上了十一国庆，估计大家已经对本周末没什么概念了，但是我们该做总结还是要做总结的。
-本周的主题其实是**简单但并不简单**，本周所选的题目大多是看一下就会的题目，但是大家看完本周的文章估计也发现了，二叉树的简答题目其实里面都藏了很多细节。 这些细节我都给大家展现了出来。
+本周的主题其实是**简单但并不简单**，本周所选的题目大多是看一下就会的题目，但是大家看完本周的文章估计也发现了，二叉树的简单题目其实里面都藏了很多细节。 这些细节我都给大家展现了出来。
## 周一
From c201b947a5bb4510b5c0214ec652c6300a2c6760 Mon Sep 17 00:00:00 2001
From: El nino <69737612+el-nino2020@users.noreply.github.com>
Date: Sat, 14 Jan 2023 15:06:56 +0800
Subject: [PATCH 4/4] =?UTF-8?q?Update=200112.=E8=B7=AF=E5=BE=84=E6=80=BB?=
=?UTF-8?q?=E5=92=8C.md?=
MIME-Version: 1.0
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原有的代码leetcode上无法通过编译
---
problems/0112.路径总和.md | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
index 7aa5f2a1..b1f01336 100644
--- a/problems/0112.路径总和.md
+++ b/problems/0112.路径总和.md
@@ -155,14 +155,14 @@ public:
以上代码精简之后如下：
```cpp
-class solution {
+class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
- if (root == null) return false;
+ if (!root) return false;
if (!root->left && !root->right && sum == root->val) {
return true;
}
- return haspathsum(root->left, sum - root->val) || haspathsum(root->right, sum - root->val);
+ return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}
};
```