LeetCode 7 Reverse Integer 这道题被血虐

最新推荐文章于 2021-01-09 00:40:58 发布

gx262091291

最新推荐文章于 2021-01-09 00:40:58 发布

阅读量344

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/gx262091291/article/details/46992117

LeetCode 专栏收录该内容

52 篇文章

订阅专栏

本文介绍了如何使用迭代方法在不溢出的情况下反转整数，包括处理特定边界情况的技巧，并探讨了算法在不同场景下的应用，如atoi转换。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):

Test cases had been added to test the overflow behavior.

解法：

public int reverse1(int x){
    	int px=0;
    	while(x!=0)
    	 {
    	    if(px>Integer.MAX_VALUE/10 || px<Integer.MIN_VALUE/10)return 0;
    	    px=px*10+x%10;
    	    x=x/10;
    	}
    	return px; 
    }

十分简单的算法，如果直接用暴力法整型转字符串来做，则会溢出。在限定如上述的情况下，满足OJ上的测试用例。对px迭代测试。

这个算法最精华之处在于，迭代到最大值以及最小值的十分之一的时候，可以合理判断是否越界。若越界则返回0.这个处理方法可以沿用到leetcode的