[leetcode] Python(7)--验证回文串(125)、字符串转换整数 (atoi)(8)、实现strStr()(28)、报数(38)、最长公共前缀(14)

从零开始的力扣（第七天）~

1.验证回文串

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
—————————————————————————————————————————

先在字符串中检测字母与数字，再进行翻转验证

class Solution:def isPalindrome(self, s: str) -> bool:ss = ''.join(filter(str.isalnum, str(s)))return ss[::-1].lower() == ss.lower()

下面说一下滤波器内的函数意义：

ss = ''.join(filter(str.isalnum, str(s)))

filter()函数为滤波器函数，第一个输入值为判断语句，第二个输入值为需要判断的内容，其中str.isalnum为判断输入是否为字母或数字。

注意后面的str(s)，如果不加str()做限定，那么题目中的s是unicode字符，python 2中不能编译，但是在python 3中可以不加str()。

直接在原字符串内判断，在过程中忽略掉符号与空格

class Solution(object):def isPalindrome(self, s):""":type s: str:rtype: bool"""s_length = len(s)if s_length <= 1:return Truei = 0j = s_length - 1while i <= j:while (s[i].isalpha() != True and s[i].isdigit() != True) and i < j:i = i + 1while (s[j].isalpha() != True and s[j].isdigit() != True) and j > i:j = j - 1if s[i].lower() != s[j].lower():return Falseelse:i = i + 1j = j - 1return True

其中isalpha()为判断是否为字母，isdigit()判断是否为数字。

2.字符串转换整数 (atoi)

请你来实现一个 atoi 函数，使其能将字符串转换成整数。

首先，该函数会根据需要丢弃无用的开头空格字符，直到寻找到第一个非空格的字符为止。

当我们寻找到的第一个非空字符为正或者负号时，则将该符号与之后面尽可能多的连续数字组合起来，作为该整数的正负号；假如第一个非空字符是数字，则直接将其与之后连续的数字字符组合起来，形成整数。

该字符串除了有效的整数部分之后也可能会存在多余的字符，这些字符可以被忽略，它们对于函数不应该造成影响。

注意：假如该字符串中的第一个非空格字符不是一个有效整数字符、字符串为空或字符串仅包含空白字符时，则你的函数不需要进行转换。

在任何情况下，若函数不能进行有效的转换时，请返回 0。

说明：

假设我们的环境只能存储 32 位大小的有符号整数，那么其数值范围为 [?231, 231 ? 1]。如果数值超过这个范围，请返回 INT_MAX (2³¹ ? 1) 或 INT_MIN (?2³¹) 。

示例 1:
输入: “42”
输出: 42

示例 2:
输入: " -42"
输出: -42
解释: 第一个非空白字符为 ‘-’, 它是一个负号。我们尽可能将负号与后面所有连续出现的数字组合起来，最后得到 -42 。

示例 3:
输入: “4193 with words”
输出: 4193
解释: 转换截止于数字 ‘3’ ，因为它的下一个字符不为数字。

示例 4:
输入: “words and 987”
输出: 0
解释: 第一个非空字符是 ‘w’, 但它不是数字或正、负号，因此无法执行有效的转换。

示例 5:
输入: “-91283472332”
输出: -2147483648
解释: 数字 “-91283472332” 超过 32 位有符号整数范围，因此返回 INT_MIN (?2³¹) 。
—————————————————————————————————————————

通过遍历所有情况，得出所有情况的结果

class Solution(object):def myAtoi(self, str):""":type str: str:rtype: int"""List = list(str.lstrip())num = []neg = []i = 0n = 1if not List:return 0if List[0].isdigit():while i < len(List) and List[i].isdigit():num.append(List[i])i += 1else:     if List[0] == "-":neg.append(1)while n < len(List) and List[n].isdigit():num.append(List[n])n += 1elif List[0] == "+":while n < len(List) and List[n].isdigit():num.append(List[n])n += 1number = int(''.join(num) if num else 0)number = -number if neg else numberif number < - 2 ** 31:return -2147483648if number > 2 ** 31 - 1:return 2147483647return number

这题是真的折磨人，每一次输出结果就看到例子里有不符合情况的，到最后都快崩溃了?，最后找到了一个比较好的解法：

1. 先将左边空格全部去掉，使用lstrip()函数；
2. 通过判断，第一位数字只有四种情况：正、负、其他字符与数字，对数字与其他情况分别看待，有正负号时从第二位开始寻找数字，并在非数字处停止循环，记住负号时要给一个负值。

通过找寻所有正则情况，这个方法是真的简单，就是比较难操作

class Solution(object):def myAtoi(self, str):""":type str: str:rtype: int"""a = re.findall(r'^[-+]?\d+', str.strip()) if not a:return 0a = int(a[0])b = 2**31if a > b-1:return b-1elif a < -b:return -belse:return a

其中最重要的就是使用re.findall()正则化函数，它返回string中所有与pattern相匹配的全部字串，返回形式为数组：

findall(pattern, string, flags=0)

这里有一个比较完整的re.findall()函数使用介绍：https://zhuanlan.zhihu.com/p/37900841、http://funhacks.net/2016/12/27/regular_expression/

3.实现strStr()

实现 strStr() 函数。
给定一个 haystack 字符串和一个 needle 字符串，在 haystack 字符串中找出 needle 字符串出现的第一个位置 (从0开始)。如果不存在，则返回 -1。

示例 1:
输入: haystack = “hello”, needle = “ll”
输出: 2

示例 2:
输入: haystack = “aaaaa”, needle = “bba”
输出: -1

说明:
当 needle 是空字符串时，我们应当返回什么值呢？这是一个在面试中很好的问题。
对于本题而言，当 needle 是空字符串时我们应当返回 0 。这与C语言的 strstr() 以及 Java的 indexOf() 定义相符。
—————————————————————————————————————————

使用内置find()函数

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

依次循环找到haystack内符合条件的第一次位置

class Solution(object):def strStr(self, haystack, needle):""":type haystack: str:type needle: str:rtype: int"""len1 = len(haystack)len2 = len(needle)if len2 == 0:return 0for i in range(len1):if haystack[i:i+len2] == needle:return ielse:continuereturn -1

4.报数

报数序列是一个整数序列，按照其中的整数的顺序进行报数，得到下一个数。其前五项如下：

1.     1
2.     11
3.     21
4.     1211
5.     111221

1 被读作 “one 1” (“一个一”) , 即 11。
11 被读作 “two 1s” (“两个一”）, 即 21。
21 被读作 “one 2”, “one 1” （“一个二” , “一个一”) , 即 1211。

给定一个正整数 n（1 ≤ n ≤ 30），输出报数序列的第 n 项。

注意：整数顺序将表示为一个字符串。

示例 1:
输入: 1
输出: “1”

示例 2:
输入: 4
输出: “1211”
—————————————————————————————————————————

通过递归的思想，将前n - 1项依次递归，通过建立字典，把一次中连续的个数与数字在字典内保存，再输出

class Solution(object):def countAndSay(self, n):""":type n: int:rtype: str"""if n == 1:return str(1)last = self.countAndSay(n-1)List = []d = {
    }d[last[-1]] = 0for i in range(len(last) - 1):if last[i] not in d:d[last[i]] = 1else:d[last[i]] += 1if last[i] != last[i + 1]:List.append(d[last[i]])List.append(last[i])d[last[i]] = 0if d[last[-1]] != 0:List.append(d[last[-1]] + 1)List.append(last[i])else:List.append(1)List.append(last[-1])return ''.join([str(i) for i in List])

总感觉写的比较繁琐，本来想的是对列表内连续数字切片，但是实际操作比较麻烦。

来看看大神的解法?

class Solution(object):def countAndSay(self, n):""":type n: int:rtype: str"""dict = {
    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dict[n]

哈哈哈，供君一乐！

5.最长公共前缀

编写一个函数来查找字符串数组中的最长公共前缀。如果不存在公共前缀，返回空字符串 “”。

示例 1:
输入: [“flower”,“flow”,“flight”]
输出: “fl”

示例 2:
输入: [“dog”,“racecar”,“car”]
输出: “”
解释: 输入不存在公共前缀。

说明:
所有输入只包含小写字母 a-z 。
—————————————————————————————————————————

最初的想法是找出列表内最短的字符串作为比较字符串，通过之后的字符串与初字符串取交集得到更新字符串，这样循环

class Solution(object):def longestCommonPrefix(self, strs):""":type strs: List[str]:rtype: str"""if not strs:return ""new_list = [len(d) for d in strs]index = new_list.index(min(new_list))temp = strs[index]for i in range(len(strs)):for j,x in enumerate(temp):if x != strs[i][j]:temp = temp[:j]return temp

但是这种方法遍历时间比较长，时间复杂度较高

比较奇特的方法，通过比较所有的字符串ASCII码找出最小与最大的，查看两者的交集就可以

class Solution(object):def longestCommonPrefix(self, strs):""":type strs: List[str]:rtype: str"""if not strs: return ""s1 = min(strs)s2 = max(strs)for i,x in enumerate(s1):if x != s2[i]:return s2[:i]return s1

非常巧妙的方法，因为ASCII码比较时是从前向后查找的，所以最大与最小的就是相差最多的。

又一个精彩操作：利用python的zip函数，把str看成list然后把输入看成二维数组，左对齐纵向压缩，然后把每项利用集合去重，之后遍历list中找到元素长度大于1之前的就是公共前缀

class Solution(object):def longestCommonPrefix(self, strs):""":type strs: List[str]:rtype: str"""if not strs: return ""ss = list(map(set, zip(*strs)))res = ""for i, x in enumerate(ss):x = list(x)if len(x) > 1:breakres = res + x[0]return res

以上就是今日经验！