一刷：6月3日到6月9日

反转字符串

 		l=0r=len(s)-1while l<r:s[l],s[r]=s[r],s[l]l+=1r-=1

反转字符串II

        res=list(s)for i in range(0,len(s),2*k):res[i:i+k]=res[i:i+k][::-1]return "".join(res)

替换数字

没有使用额外的空间

s=list(input())
l=len(s)-1
x=['r','e','b','m','u','n']
num=0
for i in s:if i.isdigit():num+=1
for i in range(num*5):s.append(0)
idx_2=len(s)-1
for idx_1 in range(l,-1,-1):if s[idx_1].isdigit():for sx in x:s[idx_2]=sxidx_2-=1else:s[idx_2]=s[idx_1]idx_2-=1print("".join(s))

翻转字符串里的单词

 		s.strip()s=s[::-1]return " ".join([x[::-1] for x in s.split()])

右旋转字符串

这么写超时了。。。

k=int(input())
s=list(input())
for i in range(k):x=s[-1]for i in range(len(s)-1,0,-1):s[i]=s[i-1]s[0]=x
print("".join(s))

这个跟上一题翻转字符串里的单词有异曲同工之妙

k=int(input())
s=list(input())
s=s[::-1]
s[0:k]=s[0:k][::-1]
s[k:]=s[k:][::-1]
print("".join(s))

实现strStr(())

KMP算法的实现

i是后缀末尾
j是前缀末尾

def getnext(self,s,next):j=0next[0]=0for i in range(1,len(s)):while j>0 and s[j]!=s[i]:j=next[j-1]if s[i]==s[j]:j+=1next[i]=jreturn nextdef strStr(self, haystack, needle):next=[0]*len(needle)self.getnext(needle,next)j=0for i in range(len(haystack)):while j>0 and haystack[i]!=needle[j]:j=next[j-1]if haystack[i]==needle[j]:j+=1if j==len(needle):return i-len(needle)+1return -1

重复的子字符串

 def getnext(self,s):next=[0]*len(s)j=0next[0]=0for i in range(1,len(s)):while j>0 and s[j]!=s[i]:j=next[j-1]if s[i]==s[j]:j+=1next[i]=jreturn next[-1]  #返回最大相等公共子字符串的最大数def repeatedSubstringPattern(self, s):num=self.getnext(s)if len(s)%(len(s)-num)==0 and num!=0: #有可能返回为0,如"abac"return Trueelse:return False