题目:
编写一个函数,其作用是将输入的字符串反转过来。输入字符串以字符数组 s 的形式给出。
不要给另外的数组分配额外的空间,你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
示例 1:
输入:s = [“h”,“e”,“l”,“l”,“o”]
输出:[“o”,“l”,“l”,“e”,“h”]
示例 2:
输入:s = [“H”,“a”,“n”,“n”,“a”,“h”]
输出:[“h”,“a”,“n”,“n”,“a”,“H”]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-string
分析:
直接将首尾进行交换。
代码:
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
n = len(s)
j = n - 1
for i in range(n//2):
x = s[i]
s[i] = s[j]
s[j] = x
j -= 1
题目:
给你一个 32 位的有符号整数 x ,返回将 x 中的数字部分反转后的结果。
如果反转后整数超过 32 位的有符号整数的范围 [−231, 231 − 1] ,就返回 0。
假设环境不允许存储 64 位整数(有符号或无符号)。
示例 1:
输入:x = 123
输出:321
示例 2:
输入:x = -123
输出:-321
示例 3:
输入:x = 120
输出:21
示例 4:
输入:x = 0
输出:0
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-integer
分析:
将数字取绝对值后,转换为字符串,然后直接用切片操作将字符串倒序,然后再将其转换为整形,判断最初的值的正负,确定最后返回值的正负,最后判断是否在范围内后返回值即可。
代码:
class Solution:
def reverse(self, x: int) -> int:
y = abs(x) #取绝对值
y = str(y) #转为字符串
z = y[::-1] #倒序
z = int(z) #转为整型
if x < 0: #判断正负
z = -z
if z < (-2) ** 31 or z > 2**31 - 1: #判断是否在取值范围内
return 0
return z
题目:
给定一个字符串,找到它的第一个不重复的字符,并返回它的索引。如果不存在,则返回 -1。
示例:
s = “leetcode”
返回 0
s = “loveleetcode”
返回 2
提示:
你可以假定该字符串只包含小写字母。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/first-unique-character-in-a-string
分析:
我用了两种方式。
代码:
第一种:
class Solution:
def firstUniqChar(self, s: str) -> int:
count = collections.Counter(s)
for i, ch in enumerate(s):
if count[ch] == 1:
return i
return -1
第二种:
class Solution:
def firstUniqChar(self, s: str) -> int:
n = [[0] for i in range(26)] #二维数组
count = [] #存储只出现了一次的字符的下标
for i in range(len(s)):
x = ord(s[i]) - ord('a') #用ascll码来判断字符
n[x][0] += 1 #出现次数统计
n[x].append(i) #添加下标
for i in range(26): #找出只出现一次的字符
if n[i][0] == 1:
count.append(n[i][1])
count.sort() #下标排序
if len(count) == 0:
return -1
return count[0] #输出第一个
题目:
给定两个字符串 s 和 t ,编写一个函数来判断 t 是否是 s 的字母异位词。
注意:若 s 和 t 中每个字符出现的次数都相同,则称 s 和 t 互为字母异位词。
示例 1:
输入: s = “anagram”, t = “nagaram”
输出: true
示例 2:
输入: s = “rat”, t = “car”
输出: false
提示:
1 <= s.length, t.length <= 5 * 104
s 和 t 仅包含小写字母
分析:
Python可以直接使用函数进行解决。但是不推荐,因为属实有点赖皮了hhh。我是直接使用将字符串排序后,遍历字符串,若有不同则不是异位词,若没有则是。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/valid-anagram
代码:
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
return collections.Counter(s)==collections.Counter(t)
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
s = list(s)
t = list(t)
n = len(s)
m = len(t)
if n != m:
return False
s.sort()
t.sort()
for i in range(n):
if s[i] != t[i]:
return False
return True
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
解释:“amanaplanacanalpanama” 是回文串
示例 2:
输入: “race a car”
输出: false
解释:“raceacar” 不是回文串
提示:
1 <= s.length <= 2 * 105
字符串 s 由 ASCII 字符组成
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/valid-palindrome
分析:
不考虑字母的大小写我们可以统一为小写,然后去点字符串中的其余字符,我们可以用 ascll 码来进行过滤。然后再双指针就可以得到结果了。
代码:
class Solution:
def isPalindrome(self, s: str) -> bool:
s = s.lower()
ss = []
a = int(ord('a'))
z = int(ord('z'))
asc0 = int(ord('0'))
asc9 = int(ord('9'))
n = len(s)
for i in range(n):
x = int(ord(s[i]))
letter = x >= a and x <= z
num = x >= asc0 and x <= asc9
if letter or num:
ss.append(s[i])
m = len(ss)
if m == 0:
return True
j = m - 1
for i in range(m//2):
if ss[i] != ss[j]:
return False
j -= 1
return True
这道题我有单独写文章,见:
https://blog.csdn.net/youngwyj/article/details/122662959
题目:
实现 strStr() 函数。
给你两个字符串 haystack 和 needle ,请你在 haystack 字符串中找出 needle 字符串出现的第一个位置(下标从 0 开始)。如果不存在,则返回 -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。
示例 1:
输入:haystack = “hello”, needle = “ll”
输出:2
示例 2:
输入:haystack = “aaaaa”, needle = “bba”
输出:-1
示例 3:
输入:haystack = “”, needle = “”
输出:0
提示:
0 <= haystack.length, needle.length <= 5 * 104
haystack 和 needle 仅由小写英文字符组成
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/implement-strstr
分析:
先判断两个字符串的长度是否合法,即 haystack 长度比 neddle 长或相等。然后从 haystack 的第一个字符开始遍历,如果 haystack 的第 i 个字符与第 neddle 的第一个字符相同,则判断 haystack 后面的字符是否能组成 neddle ,若能则返回下标即可。
代码:
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
n = len(haystack) #h长度
m = len(needle) #n长度
if m == 0:
return 0
if n < m:
return -1
for i in range(n - m + 1): #遍历h,遍历的长度为h的长度减去n的长度加一,因为长度不够一定组不成n
if haystack[i] == needle[0]: #h的第i个字符是否与n的第一个字符相同
if haystack[i:i + m] == needle: #h后面的m个字符是否与n相同
return i
return -1
题目:
给定一个正整数 n ,输出外观数列的第 n 项。
「外观数列」是一个整数序列,从数字 1 开始,序列中的每一项都是对前一项的描述。
你可以将其视作是由递归公式定义的数字字符串序列:
countAndSay(1) = “1”
countAndSay(n) 是对 countAndSay(n-1) 的描述,然后转换成另一个数字字符串。
前五项如下:
1
11
21
1211
111221
第一项是数字 1
描述前一项,这个数是 1 即 “ 一 个 1 ”,记作 “11”
描述前一项,这个数是 11 即 “ 二 个 1 ” ,记作 “21”
描述前一项,这个数是 21 即 “ 一 个 2 + 一 个 1 ” ,记作 “1211”
描述前一项,这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ,记作 “111221”
要 描述 一个数字字符串,首先要将字符串分割为 最小 数量的组,每个组都由连续的最多 相同字符 组成。然后对于每个组,先描述字符的数量,然后描述字符,形成一个描述组。要将描述转换为数字字符串,先将每组中的字符数量用数字替换,再将所有描述组连接起来。
例如,数字字符串 “3322251” 的描述如下图:
示例 1:
输入:n = 1
输出:“1”
解释:这是一个基本样例。
示例 2:
输入:n = 4
输出:“1211”
解释:
countAndSay(1) = “1”
countAndSay(2) = 读 “1” = 一 个 1 = “11”
countAndSay(3) = 读 “11” = 二 个 1 = “21”
countAndSay(4) = 读 “21” = 一 个 2 + 一 个 1 = “12” + “11” = “1211”
提示:
1 <= n <= 30
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/count-and-say
分析:
这道题就是统计上个字符串中,相邻相同的数字的个数。可以直接遍历生成。
代码:
class Solution:
def countAndSay(self, n: int) -> str:
s1 = ['1']
count = 0
for i in range(n):
m = len(s1[i])
num = []
for j in range(m):
if j != m - 1:
if s1[i][j] == s1[i][j + 1]:
count += 1
else:
count += 1
num.append(str(count))
num.append(s1[i][j])
count = 0
else:
count += 1
num.append(str(count))
num.append(s1[i][j])
count = 0
a = ''.join(num)
s1.append(a)
return s1[n - 1]
当然还有,真暴力解法:
class Solution:
def countAndSay(self, n: int) -> str:
num = {
1:"1",
2:"11",
3:"21",
4:"1211",
5:"111221",
6:"312211",
7:"13112221",
8:"1113213211",
9:"31131211131221",
10:"13211311123113112211",
11:"11131221133112132113212221",
12:"3113112221232112111312211312113211",
13:"1321132132111213122112311311222113111221131221",
14:"11131221131211131231121113112221121321132132211331222113112211",
15:"311311222113111231131112132112311321322112111312211312111322212311322113212221",
16:"132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211",
17:"11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221",
18:"31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211",
19:"1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221",
20:"11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211",
21:"311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311122122111312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221131112311311121321122112132231121113122113322113111221131221",
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
24:"3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211",
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
26:"1113122113121113222123211211131211121311121321123113213221121113122123211211131221121311121312211213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121113222112131112131221121321131211132221121321132132211331121321232221123113112221131112311322311211131122211213211331121321122112133221121113122113121113222123211211131211121311121321123113111231131122112213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122113221122112133221121113122113121113222123211211131211121311121321123113213221121113122113121113222113221113122113121113222112132113213221232112111312111213322112311311222113111221221113122112132113121113222112311311222113111221132221231221132221222112112322211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112211322212322211231131122211322111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211",
27:"31131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322123211211131211121332211231131122211311122122111312211213211312111322211231131122211311123113322112111331121113112221121113122113111231133221121113122113121113222123211211131211121332211213211321322113311213211322132112311321322112111312212321121113122122211211232221123113112221131112311332111213122112311311123112111331121113122112132113311213211321222122111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311123113322113223113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331221122311311222112111312211311123113322112132113213221133122211332111213112221133211322112211213322112111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121122132112311321322112111312211312111322211213111213122112132113121113222112132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212311222122132113213221123113112221133112132123222112111312211312111322212321121113121112133221121311121312211213211312111322211213211321322123211211131211121332211213211321322113311213212312311211131122211213211331121321122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311222113111221221113122112132113121113222112132113213221133122211332111213322112132113213221132231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221",
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}
return num.get(n, None)
题目:
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:strs = [“flower”,“flow”,“flight”]
输出:“fl”
示例 2:
输入:strs = [“dog”,“racecar”,“car”]
输出:""
解释:输入不存在公共前缀。
提示:
1 <= strs.length <= 200
0 <= strs[i].length <= 200
strs[i] 仅由小写英文字母组成
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/longest-common-prefix
分析:
这道题我是直接以第一个字符串为基准,判断后面的字符串前 i 个字符是否都相同,遇到不同就结束。 按理来说找到最短的应该,我这里就没有在尝试了。
代码:
class Solution:
def longestCommonPrefix(self, strs: List[str]) -> str:
for i in range(1,len(strs[0]) + 1): #字符串的前 i 个字符
for j in range(1,len(strs)): #遍历字符串
if strs[0][:i] == strs[j][:i]:
continue
else:
return strs[0][:i - 1]
if i == len(strs[0]):
return strs[0]
return ""
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