文章作者：Tyan 博客：noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析：**Version 1，依次按顺序统计s与各个单词的字母个数，如果相同顺序（连续相等字母算顺序中的一位）的字母不相等，则不符合条件，如果s的字母个数小于3且s与单词的字母个数不等，不符合条件，如果s的字母个数大于等于3且小于单词的字母个数，不符合条件，执行到某个字符串结束，如果没都达到最后，则不符合条件，否则满足条件。Version 2只统计s一次，然后再比对。
- Version 1
class Solution:
def expressiveWords(self, s: str, words: List[str]) -> int:
count = 0
n = len(s)
for word in words:
if self.check(word, s):
count = 1
return count
def check(self, word, s):
n = len(s)
m = len(word)
if m > n:
return False
i = 0
j = 0
while i < n and j < m:
if s[i] != word[j]:
return False
t1 = 1
t2 = 1
while i < n - 1 and s[i] == s[i 1]:
t1 = 1
i = 1
while j < m - 1 and word[j] == word[j 1]:
t2 = 1
j = 1
if (t1 < 3 and t1 != t2) or t1 < t2:
return False
i = 1
j = 1
return i == n and j == m- Version 2
class Solution:
def expressiveWords(self, s: str, words: List[str]) -> int:
count = 0
n = len(s)
stat = []
i = 0
while i < n:
temp = 1
while i < n - 1 and s[i] == s[i 1]:
temp = 1
i = 1
stat.append((s[i], temp))
i = 1
for word in words:
m = len(word)
if m > n:
continue
if self.check(word, stat):
count = 1
return count
def check(self, word, stat):
m = len(word)
i = 0
j = 0
while j < m and i < len(stat):
if stat[i][0] != word[j]:
return False
temp = 1
while j < m - 1 and word[j] == word[j 1]:
temp = 1
j = 1
if (stat[i][1] < 3 and stat[i][1] != temp) or stat[i][1] < temp:
return False
i = 1
j = 1
return i == len(stat)Reference
- https://leetcode.com/problems/expressive-words/
