Reverse Integer

最新推荐文章于 2020-07-05 10:18:11 发布

原创最新推荐文章于 2020-07-05 10:18:11 发布 · 664 阅读

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本文探讨了如何实现整数反转并处理边界情况，包括处理尾部为0的整数和防止溢出的问题。

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).

public class Solution {
    public int reverse(int x) {
        // if(x==0) return 0  //处理0，可不用单独考虑
        
        boolean negative = false;
        if(x<0)  negative = true;
        
        x = Math.abs(x);
        
       // while(x%10==0){  //处理尾部的0，可不用单独考虑
        //    x/=10;
        //}
        
        int now = 0;
        while(x>0){
            now*=10;
            now+=x%10;
            x/=10;
        }
        if(now<0)  return -1; // 溢出啦！！！题目要求不用exception
        if(negative)  now = -now;
        return now;
    }
}