第三周的算法作业

Author: xiaoluome

.gitignore

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+# vscode
+.vscode/

Week_03/id_36/LeetCode_102_36.py

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+#
+# @lc app=leetcode.cn id=102 lang=python3
+#
+# [102] 二叉树的层次遍历
+#
+# https://leetcode-cn.com/problems/binary-tree-level-order-traversal/description/
+#
+# algorithms
+# Medium (56.68%)
+# Likes:    215
+# Dislikes: 0
+# Total Accepted:    29.3K
+# Total Submissions: 51.7K
+# Testcase Example:  '[3,9,20,null,null,15,7]'
+#
+# 给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
+# 
+# 例如:
+# 给定二叉树: [3,9,20,null,null,15,7],
+# 
+# ⁠   3
+# ⁠  / \
+# ⁠ 9  20
+# ⁠   /  \
+# ⁠  15   7
+# 
+# 
+# 返回其层次遍历结果：
+# 
+# [
+# ⁠ [3],
+# ⁠ [9,20],
+# ⁠ [15,7]
+# ]
+# 
+# 
+#
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+# # 递归
+# class Solution:
+#     def levelOrder(self, root: TreeNode) -> List[List[int]]:
+#         levels = []
+#         if not root:
+#             return levels
+        
+#         def helper(node, level):
+#             if len(levels) == level:
+#                 levels.append([])
+            
+#             levels[level].append(node.val)
+
+#             if node.left:
+#                 helper(node.left, level + 1)
+#             if node.right:
+#                 helper(node.right, level + 1)
+        
+#         helper(root, 0)
+#         return levels
+
+# BFS
+class Solution:
+    def levelOrder(self, root: TreeNode) -> List[List[int]]:
+        levels = []
+        if not root:
+            return levels
+        
+        level = 0
+        from collections import deque
+        queue = deque([root,])
+
+        while queue:
+            levels.append([])
+            level_length = len(queue)
+
+            for i in range(level_length):
+                node = queue.popleft()
+
+                levels[level].append(node.val)
+
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            
+            level += 1
+        
+        return levels
+

Week_03/id_36/LeetCode_104_36.py

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+#
+# @lc app=leetcode.cn id=104 lang=python3
+#
+# [104] 二叉树的最大深度
+#
+# https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/description/
+#
+# algorithms
+# Easy (69.37%)
+# Likes:    298
+# Dislikes: 0
+# Total Accepted:    52.1K
+# Total Submissions: 75.1K
+# Testcase Example:  '[3,9,20,null,null,15,7]'
+#
+# 给定一个二叉树，找出其最大深度。
+# 
+# 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
+# 
+# 说明: 叶子节点是指没有子节点的节点。
+# 
+# 示例：
+# 给定二叉树 [3,9,20,null,null,15,7]，
+# 
+# ⁠   3
+# ⁠  / \
+# ⁠ 9  20
+# ⁠   /  \
+# ⁠  15   7
+# 
+# 返回它的最大深度 3 。
+# 
+#
+# Definition for a binary tree node.
+
+
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+# # 递归
+# class Solution:
+#     def maxDepth(self, root: TreeNode) -> int:
+#         if not root:
+#             return 0
+#         return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
+
+# DFS
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        if root is None:
+            return 0
+        
+        stack = []
+        stack.append((root, 1))
+        max_depth = 0
+        while len(stack) > 0:
+            current = stack.pop()
+            root = current[0]
+            current_depth = current[1]
+            if root.left is None and root.right is None:
+                max_depth = max(max_depth, current_depth)
+            if root.left is not None:
+                stack.append((root.left, current_depth + 1))
+            if root.right is not None:
+                stack.append((root.right, current_depth + 1))
+        
+        return max_depth

Week_03/id_36/LeetCode_107_36.py

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+#
+# @lc app=leetcode.cn id=107 lang=python3
+#
+# [107] 二叉树的层次遍历 II
+#
+# https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/description/
+#
+# algorithms
+# Easy (60.90%)
+# Likes:    108
+# Dislikes: 0
+# Total Accepted:    17.6K
+# Total Submissions: 28.9K
+# Testcase Example:  '[3,9,20,null,null,15,7]'
+#
+# 给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）
+# 
+# 例如：
+# 给定二叉树 [3,9,20,null,null,15,7],
+# 
+# ⁠   3
+# ⁠  / \
+# ⁠ 9  20
+# ⁠   /  \
+# ⁠  15   7
+# 
+# 
+# 返回其自底向上的层次遍历为：
+# 
+# [
+# ⁠ [15,7],
+# ⁠ [9,20],
+# ⁠ [3]
+# ]
+# 
+# 
+#
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+# BFS
+class Solution:
+    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
+        levels = []
+        if not root:
+            return levels
+        
+        from collections import deque
+        queue = deque([root,])
+        children_list = []
+
+        while queue:
+            level_length = len(queue)
+
+            for i in range(level_length):
+                node = queue.popleft()
+
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+                
+                children_list.append(node.val)
+
+            levels.insert(0, children_list)
+            children_list = []
+        
+        return levels
+

Week_03/id_36/LeetCode_111_36.py

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+#
+# @lc app=leetcode.cn id=111 lang=python3
+#
+# [111] 二叉树的最小深度
+#
+# Definition for a binary tree node.
+
+
+class TreeNode:
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+DFS
+class Solution:
+    def minDepth(self, root: TreeNode) -> int:
+        if root is None:
+            return 0
+        
+        stack = []
+        import sys
+        stack.append((root, 1))
+        min_depth = sys.maxsize
+        while len(stack) > 0:
+            current = stack.pop()
+            root = current[0]
+            current_depth = current[1]
+            if root.left is None and root.right is None:
+                min_depth = min(min_depth, current_depth)
+            if root.left is not None:
+                stack.append((root.left, current_depth + 1))
+            if root.right is not None:
+                stack.append((root.right, current_depth + 1))
+        
+        return min_depth

Week_03/id_36/LeetCode_200_36.py

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+#
+# @lc app=leetcode.cn id=200 lang=python3
+#
+# [200] 岛屿数量
+#
+# https://leetcode-cn.com/problems/number-of-islands/description/
+#
+# algorithms
+# Medium (44.05%)
+# Likes:    171
+# Dislikes: 0
+# Total Accepted:    16.5K
+# Total Submissions: 37.4K
+# Testcase Example:  '[["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]'
+#
+# 给定一个由 '1'（陆地）和
+# '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
+# 
+# 示例 1:
+# 
+# 输入:
+# 11110
+# 11010
+# 11000
+# 00000
+# 
+# 输出: 1
+# 
+# 
+# 示例 2:
+# 
+# 输入:
+# 11000
+# 11000
+# 00100
+# 00011
+# 
+# 输出: 3
+# 
+# 
+#
+
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        def sink(i, j):
+            if 0 <= i < len(grid) and 0 <= j < len(grid[0]) and grid[i][j] == '1':
+                grid[i][j] = '0'
+                list(map(sink, (i-1, i, i+1, i), (j, j-1, j, j+1)))
+        return sum(grid[i][j] == '1' and not sink(i, j) for i in range(len(grid)) for j in range(len(grid[0])))