Leetcode 07 Integer Reverse问题

最新推荐文章于 2021-01-12 00:40:21 发布

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本文介绍了一种有效的整数反转算法，提供了两种实现方法：一种使用长整型避免溢出问题，另一种通过判断条件提前预防溢出。同时讨论了特殊输入如以0结尾的整数处理及32位整数溢出的解决方案。

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

//溢出返回0

一、用大范围的数据来存放结果

int reverse(int x) {
	long long  ret=0;
	int sign=1;
	if(x<0) 
	{
		sign=-1;
	   x=-x;
	}

	while(x)
	{
		ret=ret*10+x%10;
		x/=10;
	
	}

	 ret=sign *ret;
	 if(ret>INT_MAX || ret<INT_MIN) ret=0;
	return ret;

}

二、考虑什么时候会溢出

在计算逆转之后的整数的时候，ret=ret*10+x%10 若在这之前ret>INT_MAX/10；那么执行ret=这句后，必然会溢出。

   int reverse(int x) {
        int res = 0;
        while (x != 0) {
            if (abs(res) > INT_MAX / 10) return 0;
            res = res * 10 + x % 10;
            x /= 10;
        }
        return res;
    }

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