From cf3ff0c0034b123846be15f72a076aa2c8d9b69d Mon Sep 17 00:00:00 2001
From: "1500906945@qq.com" <1500906945@qq.com>
Date: Tue, 11 Apr 2023 22:39:33 +0800
Subject: [PATCH] =?UTF-8?q?yifan=E6=8F=90=E4=BA=A4?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 "1.1\345\210\206\346\262\273.png" | Bin 0 -> 6496 bytes
 "1.2\345\210\206\346\262\273.png" | Bin 0 -> 6222 bytes
 alg.md                            | 138 ++++++++++++++++--------------
 alg.py                            |  40 +++------
 gen.py                            |  23 ++++-
 main.cpp                          |   0
 test.py                           |  26 ++----
 7 files changed, 113 insertions(+), 114 deletions(-)
 create mode 100644 "1.1\345\210\206\346\262\273.png"
 create mode 100644 "1.2\345\210\206\346\262\273.png"
 delete mode 100644 main.cpp

diff --git "a/1.1\345\210\206\346\262\273.png" "b/1.1\345\210\206\346\262\273.png"
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diff --git a/alg.md b/alg.md
index 8a4ef77..c1ecaf6 100644
--- a/alg.md
+++ b/alg.md
@@ -1,84 +1,92 @@
-# 解题报告：构建最大二叉树
+# Leetcode240二维矩阵检索
 
-## 题目描述
+**题目描述**
 
-​	给定一个不重复的整数数组 `nums`，构建一个最大二叉树，其中根节点的值为 `nums` 中的最大值，左子树是构建在最大值左边的子数组的最大二叉树，右子树是构建在最大值右边的子数组的最大二叉树。
+在一个 m*n 的二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。
 
-## 思路分析
+示例:
+现有矩阵 matrix 如下：
+[
+ [1,  4, 7, 11, 15],
+ [2,  5, 8, 12, 19],
+ [3,  6, 9, 16, 22],
+ [10, 13, 14, 17, 24],
+ [18, 21, 23, 26, 30]
+]
+给定 target = 5，返回 true。
+给定 target = 20，返回 false。
 
-​	对于一个不重复的整数数组 `nums`，我们可以通过以下步骤构建一个最大二叉树：
+## 生成数据的思路
 
-1. 找到数组中的最大值及其索引。
-2. 创建根节点，其值为最大值。
-3. 递归地在最大值左边的子数组前缀上构建左子树。
-4. 递归地在最大值右边的子数组后缀上构建右子树。
-5. 返回构建好的最大二叉树的根节点。
+​	生成数据是算法测试中非常重要的一步，因为算法的性能往往会受到数据规模和数据分布的影响。在实际应用中，我们可能需要根据具体的场景和需求生成不同类型的数据，例如随机数、有序数列、逆序数列、重复数等等。下面是一个生成随机数列的思路：
 
-## 代码实现
+1. 首先确定矩阵的大小 m 和 n，可以随机生成一个范围内的整数作为 m 和 n，比如随机生成 2~10 之间的整数。
+2. 生成一个 m*n 的二维数组 matrix，每行和每列都是递增的，可以先生成每行的元素，再将每行添加到 matrix 中。
+3. 生成一个目标整数 target，可以随机生成一个范围内的整数作为 target，比如随机生成 1~30 之间的整数。
+4. 判断目标整数 target 是否在二维数组 matrix 中，如果存在则返回 True，否则返回 False
 
-​	下面是python代码实现，其中`TreeNode`类定义了二叉树节点的结构，`constructMaximumbinaryTree`函数接收一个整数数组 `nums`，返回该数组构建的最大二叉树的根节点。在函数内部，首先判断如果 `nums` 数组为空，则直接返回 `None`。否则，找到数组中的最大值及其索引，创建根节点，其值为最大值。然后，递归地在最大值左边的子数组前缀上构建左子树，递归地在最大值右边的子数组后缀上构建右子树。最后，返回构建好的最大二叉树的根节点。
+## 代码
 
 ```python
-# Definition for a binary tree node.
-class TreeNode:
-    def __init__(self, val=0, left=None, right=None):
-        self.val = val
-        self.left = left
-        self.right = right
-        
-def constructMaximumBinaryTree(nums):
-    # 如果数组为空，则返回None
-    if not nums:
-        return None
-        
-    max_num = max(nums)
-
-    # 找到数组中的最大值及其索引
-    idx = nums.index(max_num)
-
-    # 创建根节点，其值为最大值
-    root = TreeNode(max_num)
-
-    # 递归地在最大值左边的子数组前缀上构建左子树
-    root.left = constructMaximumBinaryTree(nums[:idx])
-
-    # 递归地在最大值右边的子数组后缀上构建右子树
-    root.right = constructMaximumBinaryTree(nums[idx+1:])
-        
-    # 返回构建好的最大二叉树的根节点
-    return root
-
+import random
+from typing import List
+def generate_matrix(m: int, n: int) -> List[List[int]]:
+    matrix = []
+    for i in range(m):
+        row = []
+        for j in range(n):
+            if i == 0 and j == 0:
+                row.append(random.randint(0, 30))
+            elif i == 0:
+                row.append(max(row[-1], random.randint(0, 30)))
+            elif j == 0:
+                col = [matrix[x][0] for x in range(i)]
+                row.append(max(col[-1], random.randint(0, 30)))
+            else:
+                col = [matrix[x][j] for x in range(i)]
+                row.append(max(row[-1], col[-1], random.randint(0, 30)))
+        matrix.append(row)
+    return matrix
 ```
 
-## 测试示例
 
-​	下面是一个测试代码的例子：
+
+## 二维矩阵检索算法的思路
+
+- 该程序首先通过 `generate_matrix` 函数生成一个大小为 `m*n` 的二维数组，其中每行和每列都是递增的。生成过程中，对于每个位置 `(i, j)`，如果它是首行或首列，则随机生成一个整数，否则选取它的左边和上边的数以及一个随机数中的最大值作为该位置的元素值。
+- 接着，程序通过 `searchMatrix` 函数在生成的二维数组中查找目标整数 `target`，该函数采用双指针法，从右上角开始遍历，如果当前元素比目标整数大，则向左移动指针，否则向下移动指针。如果遍历完整个数组-仍未找到目标整数，则返回 False。
+- 最后，程序打印出生成的二维数组和目标整数，并输出查找结果。
+
+## 代码
 
 ```python
-import alg
-nums = [3, 2, 1, 6, 0, 5]
-# 构建最大二叉树
-root = alg.constructMaximumBinaryTree(nums)
-# 前序遍历二叉树
-def preorderTraversal(root):
-    res = []
-    def helper(root):
-        # 如果树为空，则返回None
-        if not root:
-            return None
-        res.append(root.val)
-        helper(root.left)
-        helper(root.right)
-    helper(root)
-    return res
-# 打印遍历结果
-print(preorderTraversal(root))
+def searchMatrix(matrix: List[List[int]], target: int) -> bool:
+    if not matrix:
+        return False
+    row, col = 0, len(matrix[0]) - 1
+    while row < len(matrix) and col >= 0:
+        if matrix[row][col] == target:
+            return True
+        elif matrix[row][col] > target:
+            col -= 1
+        else:
+            row += 1
+    return False
+m, n = random.randint(2, 10), random.randint(2, 10)
+matrix = generate_matrix(m, n)
+print("Generated matrix:")
+for row in matrix:
+    print(row)
+target = random.randint(0, 30)
+print(f"Searching for target {target} in matrix...")
+result = searchMatrix(matrix, target)
+print(f"Result: {result}")
 ```
 
-​	这个测试代码将一个整数数组传递给 `constructMaximumBinaryTree` 函数，得到该数组构建的最大二叉树的根节点。然后，使用 `preorderTraversal` 函数进行前序遍历，并打印遍历结果。运行该测试代码将输出 `[6,3,2,1,5,0]`，即最大二叉树的前序遍历结果。
 
-## 时间复杂度分析
 
-​	对于一个长度为 `n` 的不重复的整数数组 `nums`，构建最大二叉树的时间复杂度为 $O(n\log n)$。
+## 测试结果
+
+![](C:\Users\XUZIY\Desktop\1.1分治.png)
 
-​	在构建最大二叉树的过程中，每个节点最多会被访问两次，因此时间复杂度为 $O(n)$。而在最坏情况下，数组中的每个数都会成为树的根节点，因此二叉树的深度为 $O(n)$，所以总的时间复杂度为 $O(n\log n)$。
+![](C:\Users\XUZIY\Desktop\1.2分治.png)
\ No newline at end of file
diff --git a/alg.py b/alg.py
index 35282b9..1ddd391 100644
--- a/alg.py
+++ b/alg.py
@@ -1,28 +1,12 @@
-# Definition for a binary tree node.
-class TreeNode:
-    def __init__(self, val=0, left=None, right=None):
-        self.val = val
-        self.left = left
-        self.right = right
-        
-def constructMaximumBinaryTree(nums):
-    # 如果数组为空，则返回None
-    if not nums:
-        return None
-        
-    max_num = max(nums)
-
-    # 找到数组中的最大值及其索引
-    idx = nums.index(max_num)
-
-    # 创建根节点，其值为最大值
-    root = TreeNode(max_num)
-
-    # 递归地在最大值左边的子数组前缀上构建左子树
-    root.left = constructMaximumBinaryTree(nums[:idx])
-
-    # 递归地在最大值右边的子数组后缀上构建右子树
-    root.right = constructMaximumBinaryTree(nums[idx+1:])
-        
-    # 返回构建好的最大二叉树的根节点
-    return root
+def searchMatrix(matrix: List[List[int]], target: int) -> bool:
+    if not matrix:
+        return False
+    row, col = 0, len(matrix[0]) - 1
+    while row < len(matrix) and col >= 0:
+        if matrix[row][col] == target:
+            return True
+        elif matrix[row][col] > target:
+            col -= 1
+        else:
+            row += 1
+    return False
\ No newline at end of file
diff --git a/gen.py b/gen.py
index 1c16f48..42f124a 100644
--- a/gen.py
+++ b/gen.py
@@ -1,4 +1,19 @@
-nums = [3,2,1,6,0,5]
-
-
-
+import random
+from typing import List
+def generate_matrix(m: int, n: int) -> List[List[int]]:
+    matrix = []
+    for i in range(m):
+        row = []
+        for j in range(n):
+            if i == 0 and j == 0:
+                row.append(random.randint(0, 30))
+            elif i == 0:
+                row.append(max(row[-1], random.randint(0, 30)))
+            elif j == 0:
+                col = [matrix[x][0] for x in range(i)]
+                row.append(max(col[-1], random.randint(0, 30)))
+            else:
+                col = [matrix[x][j] for x in range(i)]
+                row.append(max(row[-1], col[-1], random.randint(0, 30)))
+        matrix.append(row)
+    return matrix
\ No newline at end of file
diff --git a/main.cpp b/main.cpp
deleted file mode 100644
index e69de29..0000000
diff --git a/test.py b/test.py
index 3af3054..af4748a 100644
--- a/test.py
+++ b/test.py
@@ -1,19 +1,11 @@
 import gen
 import alg
-
-n = gen.nums
-root = alg.constructMaximumBinaryTree(n)
-# 前序遍历二叉树
-def preorderTraversal(root):
-    res = []
-    def helper(root):
-        # 如果树为空，则返回None
-        if not root:
-            return None
-        res.append(root.val)
-        helper(root.left)
-        helper(root.right)
-    helper(root)
-    return res
-# 输出遍历结果
-print(preorderTraversal(root))
+m, n = random.randint(2, 10), random.randint(2, 10)
+matrix = generate_matrix(m, n)
+print("Generated matrix:")
+for row in matrix:
+    print(row)
+target = random.randint(0, 30)
+print(f"Searching for target {target} in matrix...")
+result = searchMatrix(matrix, target)
+print(f"Result: {result}")
\ No newline at end of file
-- 
Gitee