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
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.
问题
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
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
最怕一生碌碌无为,还说平凡难能可贵。