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Reverse Integer

时间:2014-06-24 20:16:39      阅读:180      评论:0      收藏:0      [点我收藏+]

标签:java   leetcode   数字   

题目

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 integer‘s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

方法

首先判断正负,然后从整数的地位开始循环,转换为新数。
PS:没有考虑越界问题。
	    public int reverse(int x) {
	        int flag = 0;
	        if(x < 0){
	            flag = 1;
	            x = - x;
	        }
	        int y = 0;
	        int remainder;
	        while(x > 0){
	            remainder = x % 10;
	            x = x/10;
	            y = y * 10 + remainder;
	        }
	        if(flag == 1){
	            y = -y;
	        }
	        return y;
	    }


Reverse Integer,布布扣,bubuko.com

Reverse Integer

标签:java   leetcode   数字   

原文地址:http://blog.csdn.net/u010378705/article/details/33329751

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