代码随想录算法训练营第九天 | 字符串part02
151 翻转字符串里的单词
举个例子，源字符串为："the sky is blue "
- 移除多余空格 : "the sky is blue"
- 字符串反转："eulb si yks eht"
- 单词反转："blue is sky the"
（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)
# 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
# 将字符串拆分为单词，即转换成列表类型
words = s.split()
# 反转单词
left, right = 0, len(words) - 1
while left < right:
words[left], words[right] = words[right], words[left]
left += 1
right -= 1
# 将列表转换成字符串
return " ".join(words)
(版本三) 拆分字符串 + 反转列表
words = s.split() #type(words) --- list
words = words[::-1] # 反转单词
return ' '.join(words) #列表转换成字符串
卡码网：55 右旋转字符串
本人思路：
因为是右边的挪到了前面，所以先在前面补充几个0，然后和最后的几个换位置就可以了。
s = "abcdefg"
n = 2
s_ = [0] * n
s_.extend(s)
left = 0
right = len(s_) - n
while left < n:
s_[left] , s_[right] = s_[right] , s_[left]
left += 1
right += 1
list_ = []
for i in range(len(s_)):
if s_[i] != 0:
list_.append(s_[i])
print(''.join(list_))
答案方法
#获取输入的数字k和字符串
k = int(input())
s = input()
#通过切片反转第一段和第二段字符串
#注意：python中字符串是不可变的，所以也需要额外空间
s = s[len(s)-k:] + s[:len(s)-k]
print(s)
k = int(input())
s = input()
print(s[-k:] + s[:-k])
28 实现 strStr() （本题可以跳过）
自己做的，思路很简单，就是对比，这个对比首先对比头，然后对比尾，每次尾--
haystack = "a"
needle = "a"
n = len(needle) - 1
left = 0
right = left + n
list_ = []
print(left,right,n)
if len(needle) == 1:
for i in range(len(haystack)):
if haystack[i] == needle[0]:
print(i)
while left < right and left <= (len(haystack) - len(needle)):
while left < right and haystack[left] == needle[0] and haystack[right] == needle[n]:
right -= 1
n -= 1
print(left,right,n)
if left == right:
list_.append(left)
n = len(needle) - 1
left += 1
right = left + n
if len(list_) > 1:
print(list_[0])
else:
print(-1)
暴力法
m, n = len(haystack), len(needle)
for i in range(m):
if haystack[i:i+n] == needle:
return i
return -1
使用 index
try:
return haystack.index(needle)
except ValueError:
return -1
使用 find
return haystack.find(needle)
前缀表（-1）
def getNext(self, next, s):
j = -1
next[0] = j
for i in range(1, len(s)):
while j >= 0 and s[i] != s[j+1]:
j = next[j]
if s[i] == s[j+1]:
j += 1
next[i] = j
def strStr(self, haystack: str, needle: str) -> int:
if not needle:
return 0
next = [0] * len(needle)
self.getNext(next, needle)
j = -1
for i in range(len(haystack)):
while j >= 0 and haystack[i] != needle[j+1]:
j = next[j]
if haystack[i] == needle[j+1]:
j += 1
if j == len(needle) - 1:
return i - len(needle) + 1
return -1
前缀表（不-1）
def getNext(self, next: List[int], s: str) -> None:
j = 0
next[0] = 0
for i in range(1, len(s)):
while j > 0 and s[i] != s[j]:
j = next[j - 1]
if s[i] == s[j]:
j += 1
next[i] = j
def strStr(self, haystack: str, needle: str) -> int:
if len(needle) == 0:
return 0
next = [0] * len(needle)
self.getNext(next, needle)
j = 0
for i in range(len(haystack)):
while j > 0 and haystack[i] != needle[j]:
j = next[j - 1]
if haystack[i] == needle[j]:
j += 1
if j == len(needle):
return i - len(needle) + 1
return -1
459 重复的子字符串 （本题可以跳过）
（版本一） 前缀表 减一
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
if len(s) == 0:
return False
nxt = [0] * len(s)
self.getNext(nxt, s)
if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0:
return True
return False
def getNext(self, nxt, s):
nxt[0] = -1
j = -1
for i in range(1, len(s)):
while j >= 0 and s[i] != s[j+1]:
j = nxt[j]
if s[i] == s[j+1]:
j += 1
nxt[i] = j
return nxt
（版本二） 前缀表 不减一
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
if len(s) == 0:
return False
nxt = [0] * len(s)
self.getNext(nxt, s)
if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0:
return True
return False
def getNext(self, nxt, s):
nxt[0] = 0
j = 0
for i in range(1, len(s)):
while j > 0 and s[i] != s[j]:
j = nxt[j - 1]
if s[i] == s[j]:
j += 1
nxt[i] = j
return nxt
（版本三） 使用 find
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
n = len(s)
if n <= 1:
return False
ss = s[1:] + s[:-1]
print(ss.find(s))
return ss.find(s) != -1
（版本四） 暴力法
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
n = len(s)
if n <= 1:
return False
substr = ""
for i in range(1, n//2 + 1):
if n % i == 0:
substr = s[:i]
if substr * (n//i) == s:
return True
return False
