1. LeetCode1079題：活字印刷

你有一套活字字模 tiles，其中每個字模上都刻有一個字母 tiles[i]。返回你可以印出的非空字母序列的數目。

注意：本題中，每個活字字模只能使用一次。

示例 1：
輸入：“AAB”
輸出：8
解釋：可能的序列為 “A”, “B”, “AA”, “AB”, “BA”, “AAB”, “ABA”, “BAA”。

示例 2：
輸入：“AAABBC”
輸出：188

提示：
1 <= tiles.length <= 7
tiles 由大寫英文字母組成

2. 求字母序列數目的Python代碼

2.1 遞歸

def numTilePossibilities(tiles):result = 0for i in set(tiles):result += 1 + numTilePossibilities(tiles.replace(i, '', 1))return resultprint(numTilePossibilities('AAB')) print(numTilePossibilities('AABC')) print(numTilePossibilities('AAABBC'))

實驗結果：
8
34
188

代碼解讀：
函數numTilePossibilities是一個遞歸函數，參數tiles代表活字字模上的字母組成的字符串，返回值result代表使用tiles中的字母能夠排列出多少種字符串。

遞歸子問題：從tiles中拿出1個字母i后，使用剩余字母能夠排列出多少種字符串。該問題在代碼中表達為：numTilePossibilities(tiles.replace(i, '', 1))。

若剩余字母排列出的字符串都加一個前綴i，即可得到一批新的字符串。由此可得，以字母i為前綴的字符串(包括字母i本身)共有1 + numTilePossibilities(tiles.replace(i, '', 1))種。

通過for i in set(tiles):遍歷tiles去重后的字母集合，并使用result累計以不同字母為前綴的字符串的總數。

2.2 迭代

def numTilePossibilities(tiles):result = 0stack = [tiles]while stack:tiles = stack.pop()for i in set(tiles):result += 1stack.append(tiles.replace(i, '', 1))return resultprint(numTilePossibilities('AAB')) print(numTilePossibilities('AABC')) print(numTilePossibilities('AAABBC'))

實驗結果：
8
34
188

3. 求字母序列的Python代碼

3.1 遞歸

遞歸代碼對解空間樹的搜索是深度優先搜索(depth first search, dfs)。

def numTilePossibilities(tiles):result = []def dfs(tiles, prefix):for i in sorted(set(tiles)):result.append(prefix+i)dfs(tiles.replace(i, '', 1), prefix+i)dfs(tiles, '')return resultprint(numTilePossibilities('AAB')) print(numTilePossibilities('AABC'))

實驗結果：
[‘A’, ‘AA’, ‘AAB’, ‘AB’, ‘ABA’, ‘B’, ‘BA’, ‘BAA’]
[‘A’, ‘AA’, ‘AAB’, ‘AABC’, ‘AAC’, ‘AACB’, ‘AB’, ‘ABA’, ‘ABAC’, ‘ABC’, ‘ABCA’, ‘AC’, ‘ACA’, ‘ACAB’, ‘ACB’, ‘ACBA’, ‘B’, ‘BA’, ‘BAA’, ‘BAAC’, ‘BAC’, ‘BACA’, ‘BC’, ‘BCA’, ‘BCAA’, ‘C’, ‘CA’, ‘CAA’, ‘CAAB’, ‘CAB’, ‘CABA’, ‘CB’, ‘CBA’, ‘CBAA’]

3.2 迭代(棧，深度優先搜索)

本迭代代碼通過棧實現了對解空間樹的深度優先搜索。

def numTilePossibilities(tiles):result = []stack = [(tiles, '')]while stack:tiles, prefix = stack.pop()result.append(prefix)for i in sorted(set(tiles), reverse=True):stack.append((tiles.replace(i, '', 1), prefix+i))return result[1:]print(numTilePossibilities('AAB')) print(numTilePossibilities('AABC'))

實驗結果：
[‘A’, ‘AA’, ‘AAB’, ‘AB’, ‘ABA’, ‘B’, ‘BA’, ‘BAA’]
[‘A’, ‘AA’, ‘AAB’, ‘AABC’, ‘AAC’, ‘AACB’, ‘AB’, ‘ABA’, ‘ABAC’, ‘ABC’, ‘ABCA’, ‘AC’, ‘ACA’, ‘ACAB’, ‘ACB’, ‘ACBA’, ‘B’, ‘BA’, ‘BAA’, ‘BAAC’, ‘BAC’, ‘BACA’, ‘BC’, ‘BCA’, ‘BCAA’, ‘C’, ‘CA’, ‘CAA’, ‘CAAB’, ‘CAB’, ‘CABA’, ‘CB’, ‘CBA’, ‘CBAA’]

3.3 迭代(棧)

def numTilePossibilities(tiles):result = []stack = [(tiles, '')]while stack:tiles, prefix = stack.pop()for i in sorted(set(tiles)):result.append(prefix+i)stack.append((tiles.replace(i, '', 1), prefix+i))return resultprint(numTilePossibilities('AAB')) print(numTilePossibilities('AABC'))

實驗結果：
[‘A’, ‘B’, ‘BA’, ‘BAA’, ‘AA’, ‘AB’, ‘ABA’, ‘AAB’]
[‘A’, ‘B’, ‘C’, ‘CA’, ‘CB’, ‘CBA’, ‘CBAA’, ‘CAA’, ‘CAB’, ‘CABA’, ‘CAAB’, ‘BA’, ‘BC’, ‘BCA’, ‘BCAA’, ‘BAA’, ‘BAC’, ‘BACA’, ‘BAAC’, ‘AA’, ‘AB’, ‘AC’, ‘ACA’, ‘ACB’, ‘ACBA’, ‘ACAB’, ‘ABA’, ‘ABC’, ‘ABCA’, ‘ABAC’, ‘AAB’, ‘AAC’, ‘AACB’, ‘AABC’]

3.4 迭代(隊列)

本迭代代碼通過隊列實現了對解空間樹的廣度優先搜索(breadth first search, bfs)。

def numTilePossibilities(tiles):result = []queue = [(tiles, '')]while queue:tiles, prefix = queue.pop(0)for i in sorted(set(tiles)):result.append(prefix+i)queue.append((tiles.replace(i, '', 1), prefix+i))return resultprint(numTilePossibilities('AAB')) print(numTilePossibilities('AABC'))

實驗結果：
[‘A’, ‘B’, ‘AA’, ‘AB’, ‘BA’, ‘AAB’, ‘ABA’, ‘BAA’]
[‘A’, ‘B’, ‘C’, ‘AA’, ‘AB’, ‘AC’, ‘BA’, ‘BC’, ‘CA’, ‘CB’, ‘AAB’, ‘AAC’, ‘ABA’, ‘ABC’, ‘ACA’, ‘ACB’, ‘BAA’, ‘BAC’, ‘BCA’, ‘CAA’, ‘CAB’, ‘CBA’, ‘AABC’, ‘AACB’, ‘ABAC’, ‘ABCA’, ‘ACAB’, ‘ACBA’, ‘BAAC’, ‘BACA’, ‘BCAA’, ‘CAAB’, ‘CABA’, ‘CBAA’]

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