LeetCode探索模块初级算法字符串章节python3代码实现

声明：本人刚刚入门，算法能力欠佳，所写算法不是最优算法，只提供使用python3的读者以参考。

反转字符串

class Solution:
def reverseString(self, s):
"""
:type s: str
:rtype: str
"""
return s[::-1]

颠倒整数

class Solution:
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
if x>=0:
s = str(x)[::-1]
num = int(s)
else:
s = str(x)[:-len(str(x)):-1]
num = -int(s)
if num>2**31-1 or num<-2**31:
return 0
return num

字符串中的第一个唯一字符

class Solution:
def firstUniqChar(self, s):
"""
:type s: str
:rtype: int
"""
dic = {}
l = []
for i in range(len(s)):
if s[i] not in l:
if s[i] not in dic.keys():
dic[s[i]] = i
else:
l.append(s[i])
dic.pop(s[i])
res = list(dic.values())
if len(res) != 0:
return min(res)
else:
return -1

有效的字母异位词

class Solution:
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
from collections import Counter
if(Counter(s)==Counter(t)):
return True
else:
return False

验证回文字符串

class Solution:
def isPalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
s1 = s.lower()
l = []
for i in s1:
if 'a' <= i <= 'z' or '0' <= i <= '9':
l.append(i)
le = len(l)
for j in range(le//2):
if l[j] != l[-j-1]:
return False
return True

字符串转整数（atoi）

class Solution:
def myAtoi(self, str):
"""
:type str: str
:rtype: int
"""
str = str.lstrip(' ')
if len(str)>0:
flag = 1
if str[0] == '+':
str = str[1:]
elif str[0] == '-':
str = str[1:]
flag = -1
elif '0' <= str[0] <= '9':
pass
else:
return 0

new_s = ""
for s in str:
if '0' <= s <= '9':
new_s += s
else:
break
if len(new_s)>0:
if flag == -1:
n = -int(new_s)
if n >= -2**31:
return n
else:
return -2**31
else:
n = int(new_s)
if n <= 2**31-1:
return n
else:
return 2**31-1
else:
return 0
return 0

实现strStr()

class Solution:
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
return haystack.find(needle)

数数并说

class Solution:
def countAndSay(self, n):
"""
:type n: int
:rtype: str
"""
if n == 1:
return "1"
elif n == 2:
return "11"
elif n > 2:
pre = self.countAndSay(n-1)
string = ""
count = 1
for i in range(1,len(pre)):
if pre[i] != pre[i-1]:
string = string + str(count) + pre[i-1]
count = 1
else:
count += 1
string = string + str(count) + pre[i]
return string

最长公共前缀

class Solution:
def longestCommonPrefix(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
l = len(strs)
if l > 0:
le = len(strs[0])
for string in strs:
le_new = len(string)
if le_new < le:
le = le_new
for i in range(le):  # i表示一个字符串的第几个字符
for j in range(1, l):  # j表示第几个字符串
if strs[j][i] != strs[j - 1][i]:
num = i
if num > 0:
return strs[0][:num]
else:
return ""
num = le
if num > 0:
return strs[0][:num]
else:
return ""
else:
return ""