Leetcode 1160. Find Words That Can Be Formed by Characters

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

解析：Version 1，使用Counter统计字符出现的次数，然后进行比较。Version 2，直接使用字符串的count方法统计字符出现的次数进行比较。

• Version 1

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        candidates = Counter(chars)
        lengths = 0
        for word in words:
            good = True
            temp = Counter(word)
            for ch, num in temp.items():
                if ch not in candidates or temp[ch] - candidates[ch] > 0:
                    good = False
                    break
            if good:
                lengths += len(word)
        return lengths

• Version 2

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        lengths = 0
        for word in words:
            good = True
            for ch in word:
                if word.count(ch) > chars.count(ch):
                    good = False
                    break
            if good:
                lengths += len(word)
        return lengths

Reference

1. https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/