标签：buffer boolean pre 负数 lock func string solution span
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
定位：简单题
将输入的数反转输出，注意的是负数符号保持在最前，反转后的值超出int_32bit范围输出0。
简单的将传如的数分离符号后用StringBuffer进行反转，然后转化为int，此时尝试抓取NumberFormatException的错误，如有直接为0。
Java实现：
1 public class Solution { 2 public int reverse(int x) { 3 boolean isNev=false; 4 if(x<0){ 5 x=-x; 6 isNev=true; 7 } 8 StringBuffer stringBuffer=new StringBuffer(String.valueOf(x)); 9 stringBuffer=stringBuffer.reverse(); 10 int y; 11 try{ 12 y=Integer.parseInt(stringBuffer.toString()); 13 }catch (NumberFormatException e){ 14 y=0; 15 } 16 if(isNev){ 17 y=-y; 18 } 19 return y; 20 } 21 }
LeetCode 007 Reverse Integer - Java
标签：buffer boolean pre 负数 lock func string solution span
原文地址：http://www.cnblogs.com/zaritiy/p/7040434.html
