7. Reverse Integer

最新推荐文章于 2022-07-25 11:26:23 发布

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最新推荐文章于 2022-07-25 11:26:23 发布

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分类专栏： LeetCode编程

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这是一个关于反转32位整数的问题，要求在给定的32位整数范围内处理正负数，并处理反转后溢出的情况。文章提供了作者的Java实现，运行时间为47ms。

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Description

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

My code:

java:

class Solution {
    public int reverse(int x) {
        long w = (long)x;
        if(w == 0){
            return 0;
        }
        else if(w > 0){
            long result = 0;
            while(true){
                long n = w%10;
                result = result*10+n;
                w = w/10;
                if(w == 0)
                    break;
            }

            if(result > Integer.MAX_VALUE)
                return 0;

            return (int)result;
        }
        else{
            w = -w;
            long result = 0;
            while(true){
                long n = w%10;
                result = result*10+n;
                w = w/10;
                if(w == 0)
                    break;
            }

            if(result > Integer.MAX_VALUE)
                return 0;

            return (int)(-result);
        }
    }
}

Runtime: 47 ms

Other code:

java:

class Solution {
    public int reverse(int x)
    {
        int result = 0;
        while (x != 0)
        {
            int tail = x % 10;
            int newResult = result * 10 + tail;
            if ((newResult - tail) / 10 != result)
            { 
                return 0; 
            }
            result = newResult;
            x = x / 10;
    }
    return result;
    }  
}