第四周作业#026

Week_04/id_26/LeetCode_169_26.py

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+#
+# @lc app=leetcode.cn id=169 lang=python
+#
+# [169] 求众数
+#
+# https://leetcode-cn.com/problems/majority-element/description/
+#
+# algorithms
+# Easy (59.48%)
+# Likes:    261
+# Dislikes: 0
+# Total Accepted:    49.3K
+# Total Submissions: 82.6K
+# Testcase Example:  '[3,2,3]'
+#
+# 给定一个大小为 n 的数组，找到其中的众数。众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。
+#
+# 你可以假设数组是非空的，并且给定的数组总是存在众数。
+#
+# 示例 1:
+#
+# 输入: [3,2,3]
+# 输出: 3
+#
+# 示例 2:
+#
+# 输入: [2,2,1,1,1,2,2]
+# 输出: 2
+#
+#
+#
+
+
+class Solution(object):
+    def majorityElement1(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        解法1：排序
+        """
+        nums.sort()
+        return nums[len(nums)//2]
+
+    def majorityElement(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        解法2: 递归+分治
+        分治问题模板步骤：
+        1.结束条件
+        2.拆分子问题
+        3.递归处理子问题
+        4.处理子问题的结果
+        """
+        def helper(left, right):
+
+            if left == right:
+                return nums[left]
+
+            mid = (right+left) // 2
+
+            leftNum = helper(left, mid)
+            rightNum = helper(mid+1, right)
+
+            if leftNum == rightNum:
+                return leftNum
+            leftCount, rightCount = 0, 0
+            for i in range(left, right+1):
+                if nums[i] == leftNum:
+                    leftCount += 1
+                elif nums[i] == rightNum:
+                    rightCount += 1
+            return leftNum if leftCount >= rightCount else rightNum
+
+        return helper(0, len(nums)-1)
+
+
+# print(Solution().majorityElement([1, 2, 1, 3, 1, 2, 1]))

Week_04/id_26/LeetCode_208_26.py

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+#
+# @lc app=leetcode.cn id=208 lang=python
+#
+# [208] 实现 Trie (前缀树)
+#
+# https://leetcode-cn.com/problems/implement-trie-prefix-tree/description/
+#
+# algorithms
+# Medium (58.88%)
+# Likes:    77
+# Dislikes: 0
+# Total Accepted:    8K
+# Total Submissions: 13.5K
+# Testcase Example:  '["Trie","insert","search","search","startsWith","insert","search"]\n[[],["apple"],["apple"],["app"],["app"],["app"],["app"]]'
+#
+# 实现一个 Trie (前缀树)，包含 insert, search, 和 startsWith 这三个操作。
+#
+# 示例:
+#
+# Trie trie = new Trie();
+#
+# trie.insert("apple");
+# trie.search("apple");   // 返回 true
+# trie.search("app");     // 返回 false
+# trie.startsWith("app"); // 返回 true
+# trie.insert("app");
+# trie.search("app");     // 返回 true
+#
+# 说明:
+#
+#
+# 你可以假设所有的输入都是由小写字母 a-z 构成的。
+# 保证所有输入均为非空字符串。
+#
+#
+#
+
+
+class TireNode(object):
+    def __init__(self):
+        self.children = {}
+        self.end = False
+
+
+class Trie(object):
+
+    def __init__(self):
+        """
+        Initialize your data structure here.
+        """
+        self.root = TireNode()
+
+    def insert(self, word):
+        """
+        Inserts a word into the trie.
+        :type word: str
+        :rtype: None
+        """
+        node = self.root
+        for c in word:
+            if c not in node.children:
+                node.children[c] = TireNode()
+            node = node.children[c]
+        node.end = True
+
+    def search(self, word):
+        """
+        Returns if the word is in the trie.
+        :type word: str
+        :rtype: bool
+        """
+        node = self.root
+        for c in word:
+            if c not in node.children:
+                return False
+            node = node.children[c]
+        return node.end
+
+    def startsWith(self, prefix):
+        """
+        Returns if there is any word in the trie that starts with the given prefix.
+        :type prefix: str
+        :rtype: bool
+        """
+        node = self.root
+        for c in prefix:
+            if c not in node.children:
+                return False
+            node = node.children[c]
+        return True
+
+
+# Your Trie object will be instantiated and called as such:
+# trie = Trie()
+# print(trie.insert("apple"))
+# print(trie.search("apple"))
+# print(trie.search("app"))
+# print(trie.startsWith("app"))
+# print(trie.insert("app"))
+# print(trie.search("app"))
+

Week_04/id_26/LeetCode_211_26.py

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+#
+# @lc app=leetcode.cn id=211 lang=python
+#
+# [211] 添加与搜索单词 - 数据结构设计
+#
+# https://leetcode-cn.com/problems/add-and-search-word-data-structure-design/description/
+#
+# algorithms
+# Medium (38.92%)
+# Likes:    51
+# Dislikes: 0
+# Total Accepted:    2.8K
+# Total Submissions: 7.3K
+# Testcase Example:  '["WordDictionary","addWord","addWord","addWord","search","search","search","search"]\n[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]'
+#
+# 设计一个支持以下两种操作的数据结构：
+#
+# void addWord(word)
+# bool search(word)
+#
+#
+# search(word) 可以搜索文字或正则表达式字符串，字符串只包含字母 . 或 a-z 。 . 可以表示任何一个字母。
+#
+# 示例:
+#
+# addWord("bad")
+# addWord("dad")
+# addWord("mad")
+# search("pad") -> false
+# search("bad") -> true
+# search(".ad") -> true
+# search("b..") -> true
+#
+#
+# 说明:
+#
+# 你可以假设所有单词都是由小写字母 a-z 组成的。
+#
+#
+
+
+class TrieNode(object):
+    def __init__(self):
+        self.children = {}
+        self.end = False
+
+
+class WordDictionary(object):
+
+    def __init__(self):
+        """
+        Initialize your data structure here.
+        """
+        self.root = TrieNode()
+
+    def addWord(self, word):
+        """
+        Adds a word into the data structure.
+        :type word: str
+        :rtype: None
+        """
+        node = self.root
+        for c in word:
+            if c not in node.children:
+                node.children[c] = TrieNode()
+            node = node.children[c]
+        node.end = True
+
+    def search(self, word):
+        """
+        Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
+        :type word: str
+        :rtype: bool
+        """
+        self.ans = False
+        self._match(self.root, word)
+        return self.ans
+
+    def _match(self, root, word):
+        if not word:
+            if root.end:
+                self.ans = root.end
+            return
+        if word[0] == '.':
+            for node in root.children.values():
+                self._match(node, word[1:])
+        if word[0] not in root.children:
+            return
+        self._match(root.children[word[0]], word[1:])

Week_04/id_26/LeetCode_240_26.py

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+#
+# @lc app=leetcode.cn id=240 lang=python
+#
+# [240] 搜索二维矩阵 II
+#
+# https://leetcode-cn.com/problems/search-a-2d-matrix-ii/description/
+#
+# algorithms
+# Medium (36.82%)
+# Likes:    111
+# Dislikes: 0
+# Total Accepted:    17.8K
+# Total Submissions: 48.2K
+# Testcase Example:  '[[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]]\n5'
+#
+# 编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性：
+#
+#
+# 每行的元素从左到右升序排列。
+# 每列的元素从上到下升序排列。
+#
+#
+# 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。
+#
+#
+
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        思路：从左下角开始，相等直接返回
+             若大于目标值，则不可能在这一列，指针向左
+             若小于目标值，则不可能在这一排，指针向上
+        时间复杂度O(m+n)
+        """
+        if not matrix or not matrix[0]:
+            return False
+        m, n = len(matrix)-1, 0
+
+        while m >= 0 and n < len(matrix[0]):
+            if matrix[m][n] == target:
+                return True
+            if matrix[m][n] < target:
+                n += 1
+            else:
+                m -= 1
+        return False
+
+
+# print(Solution().searchMatrix([[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]], 333))

Week_04/id_26/LeetCode_46_26.py

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+#
+# @lc app=leetcode.cn id=46 lang=python
+#
+# [46] 全排列
+#
+# https://leetcode-cn.com/problems/permutations/description/
+#
+# algorithms
+# Medium (69.33%)
+# Likes:    291
+# Dislikes: 0
+# Total Accepted:    28.6K
+# Total Submissions: 41.1K
+# Testcase Example:  '[1,2,3]'
+#
+# 给定一个没有重复数字的序列，返回其所有可能的全排列。
+#
+# 示例:
+#
+# 输入: [1,2,3]
+# 输出:
+# [
+# ⁠ [1,2,3],
+# ⁠ [1,3,2],
+# ⁠ [2,1,3],
+# ⁠ [2,3,1],
+# ⁠ [3,1,2],
+# ⁠ [3,2,1]
+# ]
+#
+#
+
+
+class Solution(object):
+    def permute(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        时间复杂度：O(n^2)
+        """
+        ans = []
+        n = len(nums)
+
+        def helper(path=[], invited=set()):
+            if len(path) == n:
+                return ans.append(path)
+            for j in nums:
+                if j not in invited:
+                    invited.add(j)
+                    helper(path+[j], invited)
+                    invited.remove(j)
+
+        helper()
+        return ans