7. Reverse Integer Total Accepted: 153147 Total Submissions: 644103 Difficulty: Easy Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 思路: 本题思路很简单,有多种方法:要注意的就是判断反转之后的结果是否超出了int类型的范围. 第一种是直接对10取余和除,然后每加一位,就将原先…

7. Reverse Integer Total Accepted: 153147 Total Submissions: 644103 Difficulty: Easy Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 思路: 本题思路很简单,有多种方法:要注意的就是判断反转之后的结果是否超出了int类型的范围. 第一种是直接对10取余和除,然后每加一位,就将原先…

Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 解题思路:将数字翻转并不难,可以转成String类型翻转,也可以逐位翻转,本题涉及到的主要是边界和溢出问题,使用Long或者BigInteger即可解决. 题目不难: JAVA实现如下: public class Solution { static public int reverse(int x) { if(x=…

Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -123 Output: -321 Example 3: Input: 120 Output: 21 Note:Assume we are dealing with an environment which could only hold integers within t…

public class Solution { public int reverse(int x) { long rev=0; while(x!=0){ rev = rev*10+x%10; x=x/10; if(rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE) return 0; } return (int) rev; } }…

Given a 32-bit signed integer, reverse digits of an integer.Example 1:Input: 123Output:  321Example 2:Input: -123Output: -321Example 3:Input: 120Output: 21Note:Assume we are dealing with an environment which could only hold integers within the 32-bit…

题目链接:https://leetcode.com/problems/reverse-integer/description/ Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -123 Output: -321 Example 3: Input: 120 Output: 21 Note:Assume we are dea…

六月箴言 万物之中,希望最美:最美之物,永不凋零.—— 斯蒂芬·金 第二周算法记录 007 -- Reverse Integer (整数反转) 题干英文版: Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123Output: 321 Example 2: Input: -123Output: -321 Example 3: Input: 120Output: 21 Note:Assum…

我现在在做一个叫<leetbook>的开源书项目,把解题思路都同步更新到github上了,需要的同学可以去看看 书的地址:https://hk029.gitbooks.io/leetbook/ 007. Reverse Integer[E]——处理溢出的技巧 题目 Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 思路 这题完全没丝毫的难度,任何人几分钟都可以写…

今天做了下LeetCode上面字符串倒序的题目,突然想Python中字符串倒序都有哪些方法,于是网上查了下,居然有这么多种方法: 个人觉得,第二种方法是最容易想到的,因为List中的reverse方法比较常用,做LeetCode题目7. Reverse Integer也用了这种方法,程序耗时65ms #字符串的反转 #用到for循环的步长参数,从大到小循环,到0为止 def reverse1 (s): rt = '' for i in range(len(s)-1, -1, -1): rt +=…