LeetCode | 1160. Find Words That Can Be Formed by Characters拼写单词【Python】

LeetCode 1160. Find Words That Can Be Formed by Characters拼写单词【Easy】【Python】【字符串】

Problem

LeetCode

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. 1 <= words.length <= 1000
2. 1 <= words[i].length, chars.length <= 100
3. All strings contain lowercase English letters only.

问题

力扣

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. 1 <= words.length <= 1000
2. 1 <= words[i].length, chars.length <= 100
3. 所有字符串中都仅包含小写英文字母

思路

字符串

解法一

用 collections，代码风格比较 pythonic。

Python3代码

from typing import List

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        # solution one
        import collections
        res = 0
        cnt = collections.Counter(chars)
        for word in words:
            c = collections.Counter(word)
            if all([c[i] <= cnt[i] for i in c]):
                res += len(word)
        return res

解法二

判断 word 中各个字符个数是否 <= chars 中这些字符个数。

Python3代码

from typing import List

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        # solution two
        res = 0
        for word in words:
            n = len(word)
            cnt = 0
            for i in word:
                # word 中字符 i 个数 <= chars 中字符 i 个数
                if word.count(i) <= chars.count(i):
                    cnt += 1
                else:
                    break
            # word 可以由 chars 拼出
            if cnt == n:
                res += cnt
        return res

代码地址

GitHub链接

来源：https://www.cnblogs.com/wonz/p/12511157.html