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Leetcode 数 Reverse Integer

时间:2014-05-14 00:59:13      阅读:369      评论:0      收藏:0      [点我收藏+]

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本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie


Reverse Integer

 Total Accepted: 17472 Total Submissions: 43938

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?

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).


题意:反转一个给定的数
思路:
x % 10 取末位
x / 10 去掉末位
扩展:
下面的方法是取首位和去掉首位的

int n = 1;// 得到pow(10,x的倍数)
while(x / n >= 10){
    n *= 10;
}     
x / n 取首位 
x % n 去掉首位

还可以扩展为取前k位,去掉前k位,可要把上面while循环里的10对应改变就可以了


复杂度:时间O(log n),空间O(1)

相关题目: Palindrome Number

class Solution {
public:
	int reverse(int x) {


		int reverse_x = 0;
		while(x){
			reverse_x = reverse_x * 10 + x % 10;
			x /= 10;
		}
		return reverse_x ;
	}
};


Leetcode 数 Reverse Integer,布布扣,bubuko.com

Leetcode 数 Reverse Integer

标签:des   style   blog   class   code   c   

原文地址:http://blog.csdn.net/zhengsenlie/article/details/25630865

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