[LeetCode] Reverse Integer 翻转整数】的更多相关文章

Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought throu…
Reverse digits of an integer. Returns 0 when the reversed integer overflows (signed 32-bit integer). Have you met this question in a real interview?     Example Given x = 123, return 321 Given x = -123, return -321 LeetCode上的原题,请参见我之前的博客Reverse Integ…
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 store integers within…
Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought throug…
Reverse Pairs 翻转对 题意 计算数组里面下标i小于j,但是i的值要大于j的值的两倍的搭配的个数(也就是可能会有多种搭配):网址 做法 这道题显然是不允许使用最简单的方法:两次循环,逐次进行判断,这样做的时间复杂度就是O(n^2),OJ无法通过,需要考虑另外的实现方式: class Solution(object): def reversePairs(self, nums): """ :type nums: List[int] :rtype: int "…
Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 click to show spoilers. SOLUTION 1: 注意越界后返回0.先用long来计算,然后,把越界的处理掉. public class Solution { public int reverse(int x) { long ret = 0; while (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 题意: 给定一个10进制整数,翻转它. Solution1: directly do the simulation. Two tricky parts to be hand…
问题: Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return –321   Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! If the in…
题目等级:Easy 题目描述: 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    题意:给出一个 32 位的有符号整数,你需要将这个整数中每位上的数字进行反转. 解题思路:   本题很简单,我们给出以下两种方法.  …
描述: 给定32位整数,反转,如321转成123. 解决: 关键是溢出检测: int reverse(int x) { ; int temp; while (x) { temp = ret * + x % ; != ret) ; ret = temp; x /= ; } return ret; } 看了下其他答案,还有一些思路: 先声明个long,看最后是否溢出,这样只有long是64位时可以,或者用int64_t. 还有先转字符串反转再转数字的.…