leetcode 1160. Find Words That Can Be Formed by Character 解法 Python

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本文解析了一道LeetCode题目，目标是在给定字符集chars中找出所有字符串words能组成的字符串总长度。通过遍历每个单词并检查其字符是否在chars中且数量足够，实现了解题策略。文章提供了Python代码示例。

一.问题描述

ou 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.

二.解题思路

对于每个word，遍历，对每个char判断该char是否在chars里和该char在word中的数量是否小于或等于在chars中的数量。

都满足的话迭代下一个char，如果全部ch都满足，则该word符合。

可以增加一个剪枝操作，如果word的长度大于chars的长度，那么一定不符合。

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三.源码

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        len_chars=len(chars)
        rst=0
        for word in words:
            if len(word)>len_chars:continue
            Flag=1
            for ch in word:
                if  ch not in chars or chars.count(ch)<word.count(ch):
                    Flag=0
                    break
            rst+=Flag*len(word)
        return rst