数字逆序 leetcode

最新推荐文章于 2025-02-10 21:22:44 发布

原创最新推荐文章于 2025-02-10 21:22:44 发布 · 730 阅读

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本文介绍了一种用于反转整数的算法实现，通过示例展示了如何处理正负整数的反转，并讨论了处理尾部0及整数溢出等问题。

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;
    }
}