Update 0151.翻转字符串里的单词.md

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jianghongcheng
2023-05-06 17:26:08 -05:00
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@ -434,134 +434,40 @@ class Solution {
```
python:
(版本一)先删除空白,然后整个反转,最后单词反转。
### 因为字符串是不可变类型所以反转单词的时候需要将其转换成列表然后通过join函数再将其转换成列表所以空间复杂度不是O(1)
```Python
class Solution:
#1.去除多余的空格
def trim_spaces(self, s):
n = len(s)
left = 0
right = n-1
while left <= right and s[left] == ' ': #去除开头的空格
left += 1
while left <= right and s[right] == ' ': #去除结尾的空格
right -= 1
tmp = []
while left <= right: #去除单词中间多余的空格
if s[left] != ' ':
tmp.append(s[left])
elif tmp[-1] != ' ': #当前位置是空格,但是相邻的上一个位置不是空格,则该空格是合理的
tmp.append(s[left])
left += 1
return tmp
#2.翻转字符数组
def reverse_string(self, nums, left, right):
while left < right:
nums[left], nums[right] = nums[right], nums[left]
left += 1
right -= 1
return None
#3.翻转每个单词
def reverse_each_word(self, nums):
start = 0
end = 0
n = len(nums)
while start < n:
while end < n and nums[end] != ' ':
end += 1
self.reverse_string(nums, start, end-1)
start = end + 1
end += 1
return None
#4.翻转字符串里的单词
def reverseWords(self, s): #测试用例:"the sky is blue"
l = self.trim_spaces(s) #输出:['t', 'h', 'e', ' ', 's', 'k', 'y', ' ', 'i', 's', ' ', 'b', 'l', 'u', 'e'
self.reverse_string(l, 0, len(l)-1) #输出:['e', 'u', 'l', 'b', ' ', 's', 'i', ' ', 'y', 'k', 's', ' ', 'e', 'h', 't']
self.reverse_each_word(l) #输出:['b', 'l', 'u', 'e', ' ', 'i', 's', ' ', 's', 'k', 'y', ' ', 't', 'h', 'e']
return ''.join(l) #输出blue is sky the
def reverseWords(self, s: str) -> str:
# 删除前后空白
s = s.strip()
# 反转整个字符串
s = s[::-1]
# 将字符串拆分为单词,并反转每个单词
s = ' '.join(word[::-1] for word in s.split())
return s
```
(版本二)使用双指针
```python
class Solution:
def reverseWords(self, s: str) -> str:
# method 1 - Rude but work & efficient method.
s_list = [i for i in s.split(" ") if len(i) > 0]
return " ".join(s_list[::-1])
# 将字符串拆分为单词,即转换成列表类型
words = s.split()
# method 2 - Carlo's idea
def trim_head_tail_space(ss: str):
p = 0
while p < len(ss) and ss[p] == " ":
p += 1
return ss[p:]
# Trim the head and tail space
s = trim_head_tail_space(s)
s = trim_head_tail_space(s[::-1])[::-1]
pf, ps, s = 0, 0, s[::-1] # Reverse the string.
while pf < len(s):
if s[pf] == " ":
# Will not excede. Because we have clean the tail space.
if s[pf] == s[pf + 1]:
s = s[:pf] + s[pf + 1:]
continue
else:
s = s[:ps] + s[ps: pf][::-1] + s[pf:]
ps, pf = pf + 1, pf + 2
else:
pf += 1
return s[:ps] + s[ps:][::-1] # Must do the last step, because the last word is omit though the pointers are on the correct positions,
```
```python
class Solution: # 使用双指针法移除空格
def reverseWords(self, s: str) -> str:
def removeextraspace(s):
start = 0; end = len(s)-1
while s[start]==' ':
start+=1
while s[end]==' ':
end-=1
news = list(s[start:end+1])
slow = fast = 0
while fast<len(news):
while fast>0 and news[fast]==news[fast-1]==' ':
fast+=1
news[slow]=news[fast]
slow+=1; fast+=1
#return "".join(news[:slow])
return news[:slow]
def reversestr(s):
left,right = 0,len(s)-1
news = list(s)
# 反转单词
left, right = 0, len(words) - 1
while left < right:
news[left],news[right] = news[right],news[left]
left+=1; right-=1
#return "".join(news)
return news
words[left], words[right] = words[right], words[left]
left += 1
right -= 1
news = removeextraspace(s)
news.append(' ')
fast=slow=0
#print(news)
while fast<len(news):
while news[fast]!=' ':
fast+=1
news[slow:fast] = reversestr(news[slow:fast])
# print(news[slow:fast])
fast=slow=fast+1
news2 = reversestr(news[:-1])
return ''.join(news2)
# 将列表转换成字符串
return " ".join(words)
```
Go
```go