704. Binary Search Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1. You must write an algorithm with O(log n) runtime complexity. 規則 給予一個數字 target 和一個數組 nums，如果 nums 中存在 target，則返回 target 的索引，反之返回 -1。 測試案例 Example 1: Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4 Example 2: Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1 思路 取出中間值 如果 low > high ,則代表沒找到，回傳 -1 如果中間值等於 target，命中則回傳 如果中間值大於 target，則從中間值開始往左找 如果中間值小於 target，則從中間值開始往右找 實作 def search(self, nums, target) -> int: return self....

46. Permutations Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order. 規則 全排列 測試案例 Example 1: Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] Example 2: Input: nums = [0,1] Output: [[0,1],[1,0]] Example 3: Input: nums = [1] Output: [[1]] 思路 列出所有可能性 觀察底下為 swap 後即可 合併為同一陣列輸出 深度優先 實作 result = [] # 初始化 result # {0, 1, 2} #{1} -> 2 #{2} -> 2 #{3} -> 2 for i in range(start, len(nums)): # swap 值交換，以便生成不同的排列。這是排列算法的關鍵部分，通過不斷交換元素，生成不同的排列。 # 外層開始 nums[0] = 1 # 外層開始 nums[1] = 2 # 外層開始 nums[2] = 3 nums[start], nums[i] = nums[i], nums[start] result += self....

383. Ransom Note Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise. Each letter in magazine can only be used once in ransomNote. 規則 給予 ransomNote(target) 和 magazine(dict) 拼字，如果拼得出來=True，反之為 False。 測試案例 Example 1: Input: ransomNote = "a", magazine = "b" Output: false Example 2: Input: ransomNote = "aa", magazine = "ab" Output: false Example 3: Input: ransomNote = "aa", magazine = "aab" Output: true 思路 根據規則每個字只能用一次，如果重複使用則返回 False 創建 hashmap 儲存 magazine values loop 依序遞減 解法 def canConstruct(self, ransomNote: str, magazine: str): values = {} for c in magazine: if c not in values: # 不存在 hashmap 內 values[c] = 1 # 建立 else: values[c] += 1 # 有則+1 for c in ransomNote: if c not in values or values[c] == 0: # 不存在或已經用完 return False else: values[c] -= 1 # 有則-1 return True 使用內建 defaultdict 來處理 hashmap，解法大同小異，但速度較快且代碼更簡潔。...

Global search 搜尋條件 以什麼欄位搜尋 // 單獨 protected static ?string $recordTitleAttribute = 'title'; // 多個 public static function getGloballySearchableAttributes(): array { return ['title', 'slug', 'author.name', 'category.name']; } 客製化搜尋結果欄位顯示，舉例來說 title = Yish 前面想要添加 author.name: public static function getGlobalSearchResultTitle(Model $record): string { return $record->author->name .':'. $record->name; } 添加更多內容在搜尋結果欄位內： public static function getGlobalSearchResultDetails(Model $record): array { return [ 'Author' => $record->author->name, 'Category' => $record->category->name, ]; } 使用何種 query 來作搜尋，這邊可以看到我們必須先 eager load author 和 category:...

1. Two Sum Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order. 規則 給一組陣列以及 target，回傳某兩數 index 相加等於 target 數。 測試案例 Example 1: Input: nums = [2,7,11,15], target = 9 Output: [0,1] Explanation: Because nums[0] + nums[1] == 9, we return [0, 1]....

34. Find First and Last Position of Element in Sorted Array Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1]. You must write an algorithm with O(log n) runtime complexity. 規則 尋找已排序過的陣列找到 first 和 last target 數 測試案例 Example 1: Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4] Example 2: Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1] Example 3: Input: nums = [], target = 0 Output: [-1,-1] 思路 透過已排序過的輸入，透過二元搜尋法切割查找 找到 first 和 last，回傳 index 如果沒找到則回傳 -1 實作 nums = [1, 3, 3, 5, 7, 8, 9, 9, 9, 15] target = 9 print(Solution()....