Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 题目大意 反转int的数字. 难度系数: 容易 实现 易错点是没有做溢出检查.后来我看了下网上的答案,写的比我好很多(好惭愧).所以这题就给出个网上的答案吧. int reverse(int x) { int y=0; int n; while( x != 0){ n = x%10; if (y > INT_MAX
一:Reverse Integer 题目: Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 链接:https://leetcode.com/problems/reverse-integer/ 分析:这题通过不断取余将余数存放在一个vector中,然后乘以相应的10^i次方相加即可,这里主要考虑是否overflow,因此将result设为long long int
Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 若反转的数溢出,直接返回0 可以用计算结果来判断溢出,也可以用因数来判断 Java代码实现: 1 public class ReverseInteger { 2 public static int reverseInt(int x){ 3 if (x == 0) { 4 return
Reverse Integer 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 alread
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 Integer.
题目来源:https://leetcode.com/problems/reverse-integer/ Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 解题思路: 其实这道题看起来非常简单,要实现也是几行代码的事.但是有个小问题容易被忽略,就是边界问题.什么意思呢?如果我们输入的整数超出了int的表达范围,这个问题要怎么解决呢? 用比int更大的数据类型存储我们转