From e93f00aef737dad56bf4656a02c17cb66e214fcc Mon Sep 17 00:00:00 2001 From: ForsakenDelusion <144082461+ForsakenDelusion@users.noreply.github.com> Date: Sat, 21 Dec 2024 18:06:53 +0800 Subject: [PATCH 1/6] =?UTF-8?q?0235.=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96?= =?UTF-8?q?=E5=85=88=E4=B8=80=E5=A4=84=E9=94=99=E5=88=AB=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0235.二叉搜索树的最近公共祖先.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index 192bb031..3911261a 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -99,7 +99,7 @@ if (cur == NULL) return cur; * 确定单层递归的逻辑 -在遍历二叉搜索树的时候就是寻找区间[p->val, q->val]（注意这里是左闭又闭） +在遍历二叉搜索树的时候就是寻找区间[p->val, q->val]（注意这里是左闭右闭） 那么如果 cur->val 大于 p->val，同时 cur->val 大于q->val，那么就应该向左遍历（说明目标区间在左子树上）。 From 173ebe75726cf41260ccc8824c73fcb78dfdcd8f Mon Sep 17 00:00:00 2001 From: "Zhen (Evan) Wang" <273509239@qq.com> Date: Wed, 25 Dec 2024 11:14:51 +0800 Subject: [PATCH 2/6] =?UTF-8?q?=E6=9B=B4=E6=96=B00059.=E8=9E=BA=E6=97=8B?= =?UTF-8?q?=E7=9F=A9=E9=98=B5II=E7=9A=84C#=E7=89=88=E6=9C=AC=E4=BF=9D?= =?UTF-8?q?=E8=AF=81=E4=B8=8EC++=E7=89=88=E6=9C=AC=E7=9A=84=E4=BB=A3?= =?UTF-8?q?=E7=A0=81=E4=BF=9D=E6=8C=81=E4=B8=80=E8=87=B4=E6=80=A7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0059.螺旋矩阵II.md | 77 +++++++++++++++++++++++++-------- 1 file changed, 58 insertions(+), 19 deletions(-) diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md index c59ec033..94966126 100644 --- a/problems/0059.螺旋矩阵II.md +++ b/problems/0059.螺旋矩阵II.md @@ -715,26 +715,65 @@ object Solution { ### C#： ```csharp -public class Solution { - public int[][] GenerateMatrix(int n) { - int[][] answer = new int[n][]; - for(int i = 0; i < n; i++) - answer[i] = new int[n]; - int start = 0; - int end = n - 1; - int tmp = 1; - while(tmp < n * n) - { - for(int i = start; i < end; i++) answer[start][i] = tmp++; - for(int i = start; i < end; i++) answer[i][end] = tmp++; - for(int i = end; i > start; i--) answer[end][i] = tmp++; - for(int i = end; i > start; i--) answer[i][start] = tmp++; - start++; - end--; - } - if(n % 2 == 1) answer[n / 2][n / 2] = tmp; - return answer; +public int[][] GenerateMatrix(int n) +{ + // 参考Carl的代码随想录里面C++的思路 + // https://www.programmercarl.com/0059.%E8%9E%BA%E6%97%8B%E7%9F%A9%E9%98%B5II.html#%E6%80%9D%E8%B7%AF + int startX = 0, startY = 0; // 定义每循环一个圈的起始位置 + int loop = n / 2; // 每个圈循环几次，例如n为奇数3，那么loop = 1 只是循环一圈，矩阵中间的值需要单独处理 + int count = 1; // 用来给矩阵每个空格赋值 + int mid = n / 2; // 矩阵中间的位置，例如：n为3， 中间的位置就是(1，1)，n为5，中间位置为(2, 2) + int offset = 1;// 需要控制每一条边遍历的长度，每次循环右边界收缩一位 + + // 构建result二维数组 + int[][] result = new int[n][]; + for (int k = 0; k < n; k++) + { + result[k] = new int[n]; } + + int i = 0, j = 0; // [i,j] + while (loop > 0) + { + i = startX; + j = startY; + // 四个For循环模拟转一圈 + // 第一排，从左往右遍历，不取最右侧的值(左闭右开) + for (; j < n - offset; j++) + { + result[i][j] = count++; + } + // 右侧的第一列，从上往下遍历，不取最下面的值(左闭右开) + for (; i < n - offset; i++) + { + result[i][j] = count++; + } + + // 最下面的第一行，从右往左遍历，不取最左侧的值(左闭右开) + for (; j > startY; j--) + { + result[i][j] = count++; + } + + // 左侧第一列，从下往上遍历，不取最左侧的值(左闭右开) + for (; i > startX; i--) + { + result[i][j] = count++; + } + // 第二圈开始的时候，起始位置要各自加1， 例如：第一圈起始位置是(0, 0)，第二圈起始位置是(1, 1) + startX++; + startY++; + + // offset 控制每一圈里每一条边遍历的长度 + offset++; + loop--; + } + if (n % 2 == 1) + { + // n 为奇数 + result[mid][mid] = count; + } + return result; } ``` From a7ad0cd812649e0cf1928d9c4e54bc4369588a19 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E2=80=98windscape=E2=80=99?= <2462269317@qq.com> Date: Sat, 11 Jan 2025 19:18:46 +0800 Subject: [PATCH 3/6] =?UTF-8?q?=E4=BF=AE=E6=94=B9leetcode-master\problems\?= =?UTF-8?q?kamacoder\0044.=E5=BC=80=E5=8F=91=E5=95=86=E8=B4=AD=E4=B9=B0?= =?UTF-8?q?=E5=9C=9F=E5=9C=B0.md=20Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/kamacoder/0044.开发商购买土地.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/problems/kamacoder/0044.开发商购买土地.md b/problems/kamacoder/0044.开发商购买土地.md index 739e2cad..efad56da 100644 --- a/problems/kamacoder/0044.开发商购买土地.md +++ b/problems/kamacoder/0044.开发商购买土地.md @@ -212,13 +212,14 @@ public class Main { int horizontalCut = 0; for (int i = 0; i < n; i++) { horizontalCut += horizontal[i]; - result = Math.min(result, Math.abs(sum - 2 * horizontalCut)); + result = Math.min(result, Math.abs((sum - horizontalCut) - horizontalCut)); + // 更新result。其中，horizontalCut表示前i行的和，sum - horizontalCut表示剩下的和，作差、取绝对值，得到题目需要的“A和B各自的子区域内的土地总价值之差”。下同。 } int verticalCut = 0; for (int j = 0; j < m; j++) { verticalCut += vertical[j]; - result = Math.min(result, Math.abs(sum - 2 * verticalCut)); + result = Math.min(result, Math.abs((sum - verticalCut) - verticalCut)); } System.out.println(result); From cb5b6e524197e1e2f66bcd3b157b37bfbc726803 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E2=80=98windscape=E2=80=99?= <2462269317@qq.com> Date: Sun, 12 Jan 2025 11:54:23 +0800 Subject: [PATCH 4/6] =?UTF-8?q?=E4=BF=AE=E6=94=B9leetcode-master\problems\?= =?UTF-8?q?0349.=E4=B8=A4=E4=B8=AA=E6=95=B0=E7=BB=84=E7=9A=84=E4=BA=A4?= =?UTF-8?q?=E9=9B=86.md=20Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0349.两个数组的交集.md | 15 +++++++++++++-- 1 file changed, 13 insertions(+), 2 deletions(-) diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 77dfc50a..93fa0931 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -123,6 +123,9 @@ public: ### Java： 版本一：使用HashSet ```Java +// 时间复杂度O(n+m+k) 空间复杂度O(n+k) +// 其中n是数组nums1的长度，m是数组nums2的长度，k是交集元素的个数 + import java.util.HashSet; import java.util.Set; @@ -145,8 +148,15 @@ class Solution { } //方法1：将结果集合转为数组 - - return resSet.stream().mapToInt(x -> x).toArray(); + return res.stream().mapToInt(Integer::intValue).toArray(); + /** + * 将 Set 转换为 int[] 数组： + * 1. stream() : Collection 接口的方法，将集合转换为 Stream + * 2. mapToInt(Integer::intValue) : + * - 中间操作，将 Stream 转换为 IntStream + * - 使用方法引用 Integer::intValue，将 Integer 对象拆箱为 int 基本类型 + * 3. toArray() : 终端操作，将 IntStream 转换为 int[] 数组。 + */ //方法2：另外申请一个数组存放setRes中的元素,最后返回数组 int[] arr = new int[resSet.size()]; @@ -538,3 +548,4 @@ end + From 059b6464c0b8968bd139764988ecbc652eebdabe Mon Sep 17 00:00:00 2001 From: "Zhen (Evan) Wang" <273509239@qq.com> Date: Tue, 14 Jan 2025 10:36:56 +0800 Subject: [PATCH 5/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?--199.=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84=E5=8F=B3=E8=A7=86?= =?UTF-8?q?=E5=9B=BE=20=20C#=20=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0102.二叉树的层序遍历中的199.二叉树的右视图 C# 代码 --- problems/0102.二叉树的层序遍历.md | 41 +++++++++++++++++++++++ 1 file changed, 41 insertions(+) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 98e1e98a..0c3b0f8c 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -1231,6 +1231,47 @@ impl Solution { } ``` +#### C#: + +```C# 199.二叉树的右视图 +public class Solution +{ + public IList RightSideView(TreeNode root) + { + var result = new List(); + Queue queue = new(); + + if (root != null) + { + queue.Enqueue(root); + } + while (queue.Count > 0) + { + int count = queue.Count; + int lastValue = count - 1; + for (int i = 0; i < count; i++) + { + var currentNode = queue.Dequeue(); + if (i == lastValue) + { + result.Add(currentNode.val); + } + + // lastValue == i == count -1 : left 先于 right 进入Queue + if (currentNode.left != null) queue.Enqueue(currentNode.left); + if (currentNode.right != null) queue.Enqueue(currentNode.right); + + //// lastValue == i == 0: right 先于 left 进入Queue + // if(currentNode.right !=null ) queue.Enqueue(currentNode.right); + // if(currentNode.left !=null ) queue.Enqueue(currentNode.left); + } + } + + return result; + } +} +``` + ## 637.二叉树的层平均值 [力扣题目链接](https://leetcode.cn/problems/average-of-levels-in-binary-tree/) From 2bcd88a096d694b295a855062339579458c0769f Mon Sep 17 00:00:00 2001 From: "Zhen (Evan) Wang" <273509239@qq.com> Date: Tue, 14 Jan 2025 11:18:41 +0800 Subject: [PATCH 6/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?-=20637.=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=B9=B3?= =?UTF-8?q?=E5=9D=87=E5=80=BC=20C#=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 637.二叉树的层平均值 C# 版本 --- problems/0102.二叉树的层序遍历.md | 29 +++++++++++++++++++++++ 1 file changed, 29 insertions(+) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 0c3b0f8c..ce53e49a 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -1599,6 +1599,35 @@ impl Solution { } ``` +#### C#: + +```C# 二叉树的层平均值 +public class Solution { + public IList AverageOfLevels(TreeNode root) { + var result= new List(); + Queue queue = new(); + if(root !=null) queue.Enqueue(root); + + while (queue.Count > 0) + { + int count = queue.Count; + double value=0; + for (int i = 0; i < count; i++) + { + var curentNode=queue.Dequeue(); + value += curentNode.val; + if (curentNode.left!=null) queue.Enqueue(curentNode.left); + if (curentNode.right!=null) queue.Enqueue(curentNode.right); + } + result.Add(value/count); + } + + return result; + } +} + +``` + ## 429.N叉树的层序遍历 [力扣题目链接](https://leetcode.cn/problems/n-ary-tree-level-order-traversal/)