解法1:字符串反转 ## 解法一:转换为字符串,要考虑溢出 ## 特殊的案例:< 2**31-1 class Solution: def reverse(self, x: int) -> int: if x>=0: ans = int(str(x)[::-1]) else: ans = -int( str(abs(x))[::-1] ) ## 不用用 -x? ## 考虑溢出 return ans if ans<2**31-1...
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflo...[LeetCode] 7. Reverse Integer* 原题链接: https://leetcode.com/problems/reverse-integer/ 1. 题目介绍 Given a 32-bit signed integer, reverse digits of an integer. Note...
reverse(head,tail) head.next = self.reverseKGroup(tail,k) return newhead def reverse(self,head1,tail): pre = None while head1!=tail: cur = head1.next head1.next = pre pre = head1 head1 = cur return pre 31. 下一个排列 class Solution(object): def reverse(arr,i,j): while i...
t = (1,2,3,4,5) makes a tuple t[start:end:count] will used slice a list from start to end by incrementing counter for next element by count. And using negative numbers will retrieve list from backwards. So print( t[ : : -1]) will print tuple in reverse. Here start,...
[Leetcode][python]Reverse Integer/反转整数 题目大意 反转整数123变为321,-123变为-321 注意:在32位整数范围内,并且001要成为1 假设我们的环境只能存储 32 位有符号整数,其数值范围是 [−2^31, 2^31 − 1]。根据这个假设,如果反转后的整数溢出,则返回 0。
刚接触python不久,被python的简洁强大迷倒了,在做leetcode,Reverse words in a string时,刚开始还是传统的思路想着怎么处理空格问题一直测试不通过,写的很罗嗦被师弟吐槽说你写的代码好丑,好心塞。 废话不多说直接奉上思路代码: 翻转字符串如:"Hello I am echo",输出"echo am I Hello" ...
https://leetcode.com/problems/reverse-integer/?tab=Description Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 这题如果是我想的话,肯定会想把它转换成数组然后首位往中间逼近着换位。。program cr...猜...
链接:https://leetcode-cn.com/problems/reverse-words-in-a-string python # 翻转字符串里的单词 class Solution: def reverseWords(self, s: str)->str: """ 双指针,字符串的综合操作 解题思路如下: -1.移除多余空格 -1.1 rm开头空字符,遇空left指针++,遇非空停止 ...
这是一个关于Python的LeetCode题解,题目是186号题目:反转字符串中的单词。解题思路: 1. 首先,我们需要将输入的字符串按照空格分割成单词列表。 2. 然后,我们遍历这个单词列表,对于每个单词,我们将其首字母大写,其余字母小写,然后拼接起来,形成一个新的单词。
* @链接:https://leetcode-cn.com/problems/palindrome-number/solution/9-hui-wen-shu-jiang-ti-jie-fu-zhi-dao-liu-lan-qi-k/ */varisPalindrome=function(x){if(x<0||(x%10==0&&x!=0)){returnfalse;}varreverseNumber=0;while(x>reverseNumber){reverseNumber=reverseNumber*10+x%10;x=parseInt...