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【LeetCode】Reverse Integer

时间:2014-05-03 17:35:30      阅读:345      评论:0      收藏:0      [点我收藏+]

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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).

解答

注意,若反转后的数超出整型范围,则会变成负数,还有,注意处理反转的方法。

public class Solution {
    public int reverse(int x) {
        boolean position=(x<0)?false:true;
        int result=0;
        x=Math.abs(x);
        while(x>0){
            result=result*10+x%10;
            x/=10;
        }
        if(result<0)
            return -1;
        if(!position)
            result*=-1;
        return result;
    }
}

---EOF---

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【LeetCode】Reverse Integer

标签:des   style   blog   class   code   java   

原文地址:http://blog.csdn.net/navyifanr/article/details/24905981

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