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问题描述:
Reverse digitsof an integer.
Example1: x =123, return 321
Example2: x =-123, return -321
问题分析:反转算法并不难,关键在于对溢出问题的考虑
代码:
public class Solution { public int reverse(int x) { long result = 0; long max = (long) Math.pow(2, 31) - 1;//int带符号数的最大值 do { result = result * 10 + x % 10; System.err.println(result); } while ((x /= 10) != 0); //注意溢出问题 if(result > max || result < -max) { return 0; } return (int)result; } }
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原文地址:http://blog.csdn.net/woliuyunyicai/article/details/44857223