Leetcode 809. Expressive Words

文章作者:Tyan
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1. Description

Expressive Words

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3s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
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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
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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

  1. https://leetcode.com/problems/expressive-words/
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