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描述
A substring is a contiguous (non-empty) sequence of characters within a string.
A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it.
Given a string word, return the number of vowel substrings in word.
Example 1:
Input: word = "aeiouu"
Output: 2
Explanation: The vowel substrings of word are as follows (underlined):
- "aeiouu"
- "aeiouu"
Example 2:
Input: word = "unicornarihan"
Output: 0
Explanation: Not all 5 vowels are present, so there are no vowel substrings.
Example 3:
Input: word = "cuaieuouac"
Output: 7
Explanation: The vowel substrings of word are as follows (underlined):
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
Example 4:
Input: word = "bbaeixoubb"
Output: 0
Explanation: The only substrings that contain all five vowels also contain consonants, so there are no vowel substrings.
Note:
1 <= word.length <= 100
word consists of lowercase English letters only.
解析
根据题意，给出了一个字符串单词 word ，让我们返回在 word 中有多少个 vowel substrings ，题目中也给出了 vowel substrings 的定义，那就是就是非空子字符串且恰好包含了 a、e、i、o、u 五种小写字母，但是个数不限。
先使用暴力解法，其实很简单：
- 初始化结果 reuslt 为 0
- 遍历 range(5, len(word) + 1) 中的每个元素 L ，表示当前的子字符串的长度
- 然后遍历 ange(len(word) - L + 1) 中的每个元素的 i ，表示从索引 i 开始找子字符串
- 通过内置函数 collections.Counter 计算 word[i:i + L] 中的字符，如果符合 vowel substrings 的条件，就将结果 result 加一
- 遍历结束返回 result 即可
从运行结果来看，虽然是超过 100% ，但是不代表暴力解法有效，可能是新题，提交答案的人数太少了。人要对自己的能力有清醒的认识。
解答
class Solution(object):
def countVowelSubstrings(self, word):
"""
:type word: str
:rtype: int
"""
result = 0
for L in range(5, len(word) + 1):
for i in range(len(word) - L + 1):
c = collections.Counter(word[i:i + L])
keys = c.keys()
if 'a' in keys and 'e' in keys and 'i' in keys and 'o' in keys and 'u' in keys and len(keys) == 5:
result += 1
return result
运行结果
Runtime: 2676 ms, faster than 100.00% of Python online submissions for Count Vowel Substrings of a String.
Memory Usage: 13.4 MB, less than 100.00% of Python online submissions for Count Vowel Substrings of a String.
解析
其实可以用集合函数 set 对上面的代码进行精简。
解答
class Solution(object):
def countVowelSubstrings(self, word):
"""
:type word: str
:rtype: int
"""
return sum(set(word[i:i+L]) == set('aeiou') for L in range(5, len(word)+1) for i in range(len(word)-L+1))
运行结果
Runtime: 316 ms, faster than 100.00% of Python online submissions for Count Vowel Substrings of a String.
Memory Usage: 13.5 MB, less than 100.00% of Python online submissions for Count Vowel Substrings of a String.
解析
可以使用类似滑动窗口的方法解题，思路很简单：
- 先初始化结果 result 为 0 ，原音集合 vowels 为 {'a','e','i','o','u'}
- 因为 vowel substring 最短为 5 ，所以遍历 range(len(word)-4) 中的每个元素 i ，表示子字符串的开始索引
- 如果 word[i] 在 vowel 中，则新建一个集合 s 将 word[i] 加入进去
- 然后遍历 range(i+1, len(word)) 中的每个元素 j ，表示遍历 i 索引之后的字符，如果 word[j] 不在 vowels 说明 word[i:j] 不能构成 vowel substring ，直接 break 进行之后的循环，否则将 word[j] 加入 s 中，如果 s 的长度为 5 说明可以构成 vowel substring 了，result 加一
- 上述两层循环结束返回 result
从运行结果可以看出，运算时间大幅度提升。
解答
class Solution(object):
def countVowelSubstrings(self, word):
"""
:type word: str
:rtype: int
"""
result = 0
vowels = {'a','e','i','o','u'}
for i in range(len(word)-4):
if word[i] in vowels:
s = set(word[i])
for j in range(i+1, len(word)):
if word[j] not in vowels:
break
s.add(word[j])
if len(s) == 5:
result += 1
return result
运行结果
Runtime: 52 ms, faster than 100.00% of Python online submissions for Count Vowel Substrings of a String.
Memory Usage: 13.5 MB, less than 100.00% of Python online submissions for Count Vowel Substrings of a String.
原题链接：leetcode.com/problems/co…
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