LeetCode 1160. 拼写单词 Find Words That Can Be Formed by Characters
Table of Contents
一、中文版
给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。
假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。
注意：每次拼写时，chars 中的每个字母都只能用一次。
返回词汇表 words 中你掌握的所有单词的 长度之和。
示例 1：
输入：words = ["cat","bt","hat","tree"], chars = "atach"
输出：6
解释：
可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。
示例 2：
输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr"
输出：10
解释：
可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。
提示：
1 <= words.length <= 1000
1 <= words[i].length, chars.length <= 100
所有字符串中都仅包含小写英文字母
二、英文版
You are given an array of strings words and a string chars. A string is good if it can be formed by characters from chars (each character can only be used once). Return the sum of lengths of all good strings in words. Example 1: Input: words = ["cat","bt","hat","tree"], chars = "atach" Output: 6 Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6. Example 2: Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr" Output: 10 Explanation: The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10. Note: 1 <= words.length <= 1000 1 <= words[i].length, chars.length <= 100 All strings contain lowercase English letters only. 来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/find-words-that-can-be-formed-by-characters 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
三、My answer
class Solution: def countCharacters(self, words: List[str], chars: str) -> int: chars_counter = collections.Counter(chars) res = 0 for word in words: word_counter = collections.Counter(word) for key in word_counter: if word_counter[key] > chars_counter[key]: break else: res += (len(word)) return res
四、解题报告
算法思想：words 列表中每一个 Word 中的字符个数都小于 chars 中字符的个数，则能拼成该 Word。
本题中主要用到了 collections.Counter() 用法，其比较巧妙的一点是：如果没有某个 key，则会默认这个 key 出现的个数为 0。示例如下：
此外，代码中设计到 Python 中 for - else 的用法，我在另外一篇博客中简单介绍过，如果感兴趣，请查看：
