云服务器virtual memory exhausted 创建swap分区

Posted 2020-12-19

题目描述：

给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写。

说明：本题中，我们将空字符串定义为有效的回文串。

示例 1:

输入: "A man, a plan, a canal: Panama"
输出: true

示例 2:

输入: "race a car"
输出: false

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-palindrome

代码实现：

  1 def isPalindrome(s):
  2     ‘‘‘
  3     验证是否为回文串
  4     :param str:
  5     :return:
  6     ‘‘‘
  7 
  8     code = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u",
  9             "v", "w", "x", "y", "z"]
 10     CODE = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U",
 11             "V", "W", "X", "Y", "Z"]
 12     num = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
 13     string = ""
 14     for ch in s:
 15         if ch in code:
 16             string += ch
 17         elif ch in CODE:
 18             string += code[CODE.index(ch)]
 19         elif ch in num:
 20             string += ch
 21 
 22     print("string=", string)
 23     if string.isspace() or len(string) == 0:
 24         return True
 25 
 26     left, right = 0, len(string) - 1
 27     while left < right:
 28         if string[left] == string[right]:
 29             left += 1
 30             right -= 1
 31         else:
 32             return False
 33 
 34     return True
 35 
 36 
 37 print("------测试isHuiwen--------")
 38 s = "aaaa"
 39 print(isPalindrome(s))
 40 
 41 
 42 def isPalindrome1(s):
 43     ‘‘‘
 44     反转字符串来判断
 45     :param s:
 46     :return:
 47     ‘‘‘
 48     # isalnum() 方法检测字符串是否由字母和数字组成。
 49     sgood = "".join(ch.lower() for ch in s if ch.isalnum())
 50     return sgood == sgood[::-1]
 51     # return tmps == tmps[::-1]
 52 
 53 
 54 print("------测试isHuiwen--------")
 55 s = "A man, a plan, a canal: Panama"
 56 print(isPalindrome1(s))
 57 
 58 
 59 def isPalindromel2(s):
 60     ‘‘‘
 61     双指针判断
 62     :param s:
 63     :return:
 64     ‘‘‘
 65     tmps = ""
 66     for ch in s:
 67         if ch.isalnum():
 68             tmps += ch.lower()
 69 
 70     print("tmps=", tmps)
 71     left, right = 0, len(tmps) - 1
 72     while left < right:
 73         if tmps[left] == tmps[right]:
 74             left += 1
 75             right -= 1
 76         else:
 77             return False
 78 
 79     return True
 80 
 81 
 82 print("------测试isHuiwen2--------")
 83 s = "s "
 84 print(isPalindromel2(s))
 85 
 86 
 87 def isPalinedrome3(s):
 88     left, right = 0, len(s) - 1
 89     while left < right:
 90         if s[left].isalnum() == True and s[right].isalnum() == True:
 91             if s[left].lower() == s[right].lower():
 92                 left += 1
 93                 right -= 1
 94             else:
 95                 return False
 96         elif s[left].isalnum() == True and s[right].isalnum() == False:
 97             right -= 1
 98         elif s[left].isalnum() == False and s[right].isalnum() == True:
 99             left += 1
100         else:
101             left += 1
102             right -= 1
103 
104     return True
105 
106 
107 print("------测试isHuiwen3--------")
108 s = " "
109 print(isPalinedrome3(s))

输出：

------测试isHuiwen--------
string= aaaa
True
------测试isHuiwen--------
True
------测试isHuiwen2--------
tmps= s
True
------测试isHuiwen3--------
True

总结：双指针法和字符串反转直接判断两种方法。isalnum()函数判断字符是否为数字或者字母。字符串的反转操作[::-1]