Leetcode Reverse Integer

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： LEETCODE 文章标签： algorithm leetcode easy

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本文探讨了在32位有符号整数范围内实现整数反转的方法，并提供了一种简洁高效的解决方案，通过循环处理每一位数字并检查反转后的整数是否溢出。

Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

My answer:

int abs(int x)
{
    if(x < 0)
    {
        return -x;
    }
    else
    {
        return x;
    }
}
int reverse(int x) {
    int y,a,b,count,temp;
    y = 0;
    a = abs(x);
    if(a != 0)
    {
        temp = a;
        a = temp / 10;
        b = temp % 10;
        y = y*10 + b;
    }
    else
    {
        if(y >= 0)
        {
            if(y >= 32767)
            {
                return 0;
            }
            else
            {
                return y;
            }
        }
        else
        {
            y = -y;
            if(y <= -32767)
            {
                return 0;
            }
            else
            {
                return y;
            }
        }
    }
    return 0;
}

good answer:

int reverse(int x) {
    long long val = 0;
   do
   {
       val = val * 10 + x % 10;
       x /= 10;
   } while (x);

   return (val > INT_MAX || val < INT_MIN) ? 0 : val;
}