力扣刷题2023-04-19-1——题目：916. 单词子集
题目：
给你两个字符串数组 words1 和 words2。
现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称字符串 b 是字符串 a 的 子集 。
- 例如，
"wrr"是"warrior"的子集，但不是"world"的子集。
如果对 words2 中的每一个单词 b，b 都是 a 的子集，那么我们称 words1 中的单词 a 是 通用单词 。
以数组形式返回 words1 中所有的通用单词。你可以按 任意顺序 返回答案。
示例 1：
输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"] 输出：["facebook","google","leetcode"]
示例 2：
输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["l","e"] 输出：["apple","google","leetcode"]
示例 3：
输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","oo"] 输出：["facebook","google"]
示例 4：
输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lo","eo"] 输出：["google","leetcode"]
示例 5：
输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["ec","oc","ceo"] 输出：["facebook","leetcode"]
提示：
1 <= words1.length, words2.length <= 1041 <= words1[i].length, words2[i].length <= 10words1[i]和words2[i]仅由小写英文字母组成words1中的所有字符串 互不相同
解答：
版本1：超时但是思路没问题
class Solution:
def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
output = []
for item_1 in words1:
flag_append = 1
for item_2 in words2:
for i in range(len(item_2)):
if item_2.count(item_2[i]) > item_1.count(item_2[i]):
flag_append = 0
break
if flag_append == 0:
break
if flag_append == 1:
output.append(item_1)
return output
版本2：优化计算
class Solution:
def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
# preprocess words2
max_counts = [0] * 26
for word in words2:
counts = [word.count(chr(i + 97)) for i in range(26)]
for i in range(26):
max_counts[i] = max(max_counts[i], counts[i])
# iterate through words1
output = []
for word in words1:
counts = [word.count(chr(i + 97)) for i in range(26)]
if all(counts[i] >= max_counts[i] for i in range(26)):
output.append(word)
return output
首先，我们使用一个长度为 26 的数组 max_counts 来存储 words2 中每个字符出现的最大计数。对于每个单词 word，我们计算它包含每个字符的计数，然后检查是否满足要求。如果 word 中包含 words2 中的每个字符且计数不小于该字符在 words2 中出现的最大计数，那么就将 word 添加到输出列表中。这样可以避免嵌套循环，提高算法的效率。