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描述
There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.
Given a string text of words separated by a single space (no leading or trailing spaces) and a string brokenLetters of all distinct letter keys that are broken, return the number of words in text you can fully type using this keyboard.
Example 1:
Input: text = "hello world", brokenLetters = "ad"
Output: 1
Explanation: We cannot type "world" because the 'd' key is broken.
Example 2:
Input: text = "leet code", brokenLetters = "lt"
Output: 1
Explanation: We cannot type "leet" because the 'l' and 't' keys are broken.
Example 3:
Input: text = "leet code", brokenLetters = "e"
Output: 0
Explanation: We cannot type either word because the 'e' key is broken.
Note:
- 1 <= text.length <= 10^4
- 0 <= brokenLetters.length <= 26
- text consists of words separated by a single space without any leading or trailing spaces.
- Each word only consists of lowercase English letters.
- brokenLetters consists of distinct lowercase English letters.
解析
根据题意，就是给出了一个字符串 text ，里面的单词都用一个空格分割，且前后没有空格缀，又给出了键盘上坏了的键 brokenLetters ，如果该单词恰好需要坏了的键则表示不能成功打出，求能顺利打出完整的单词的个数。思路比较简单：
- 将 text 按照空格分割成单词列表 words
- 初始化 result 为 words 的长度
- 外层循环遍历 words ，内层遍历每个单词的每个字符 c ，如果 c 存在于 brokenLetters 则 result 减一，然后直接进行下一个单词的判断
- 遍历结束得到的 result 即为结果
解答
class Solution(object):
def canBeTypedWords(self, text, brokenLetters):
"""
:type text: str
:type brokenLetters: str
:rtype: int
"""
words = text.split(' ')
result = len(words)
for word in words:
for c in word:
if c in brokenLetters:
result -= 1
break
return result
运行结果
Runtime: 20 ms, faster than 75.98% of Python online submissions for Maximum Number of Words You Can Type.
Memory Usage: 13.5 MB, less than 96.65% of Python online submissions for Maximum Number of Words You Can Type.
解析
另外，可以通过内置函数完成：
- 将字符串 brokenLetters 转换成集合 b
- 将 text 按照空格分割成单词列表 words
- 遍历每个单词，如果单词与 b 的交集为 0 ，则 result 加一
- 遍历结束 result 即为结果
解答
class Solution(object):
def canBeTypedWords(self, text, brokenLetters):
"""
:type text: str
:type brokenLetters: str
:rtype: int
"""
b = set(brokenLetters)
words = text.split()
result = 0
for t in words:
if len(b.intersection(t)) == 0 :
result += 1
return result
运行结果
Runtime: 20 ms, faster than 75.98% of Python online submissions for Maximum Number of Words You Can Type.
Memory Usage: 13.7 MB, less than 80.63% of Python online submissions for Maximum Number of Words You Can Type.
原题链接：leetcode.com/problems/ma…
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