[英]Reverse Integer leetcode: Why does overflow occur only if 7 or greater is added to INT_MAX?
I have the solution provided by Leetcode and the part that confuses me is the fact that adding 7 (or a lower value) to Integer.MAX_VALUE or adding -8 (or a lower value) to Integer.MIN_VALUE does not result in overflow or underflow respectively. 我有Leetcode提供的解决方案,而令我感到困惑的部分是,在Integer.MAX_VALUE中添加7(或更低的值)或在Integer.MIN_VALUE中添加-8(或更低的值)不会导致上溢或下溢分别。
My logic is that if you have Integer.MAX_VALUE, adding 1 will cause overflow. 我的逻辑是,如果您具有Integer.MAX_VALUE,则加1将导致溢出。 And if you have Integer.MIN_VALUE, subtracting 1 will cause underflow. 并且如果您有Integer.MIN_VALUE,则减1将导致下溢。 Where is my understanding of overflow and underflow wrong? 我对上溢和下溢的理解错在哪里?
class Solution {
public int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
rev = rev * 10 + pop;
}
return rev;
}
}
Yes, normally adding 1 to Integer.MAX_VALUE
will result in overflow, as will subtracting 1 from Integer.MIN_VALUE
. 是的,通常向Integer.MAX_VALUE
加1会导致溢出,也会从Integer.MIN_VALUE
减去1。 But that is not what is happening here. 但这不是这里发生的事情。
This code is performing integer division by 10
, which truncates any decimal portion. 此代码执行的整数除以10
,这会将任何小数部分都截断。 When dividing Integer.MAX_VALUE
( 2147483647
) by 10, the code anticipates multiplying by 10 and adding the next digit. 将Integer.MAX_VALUE
( 2147483647
)除以10时,该代码预期将乘以10并加下一位。 That quotient is 214748364
, and multiplying by 10
is 2147483640
, with the possibility of adding another 7 without overflowing. 该商为214748364
,乘以10
为2147483640
,并且可以加上另外的7而不会溢出。 Similarly on the negative side, dividing Integer.MAX_VALUE
( -2147483648
) by 10 yields -214748364
, multiplying by 10 yields -2147483640
, with the possibility of adding another -8
without overflowing. 类似地,在负方面,将Integer.MAX_VALUE
( -2147483648
)除以10将产生-214748364
,再乘以10将产生-2147483640
,并且可以加上另一个-8
而不会溢出。
This code takes into account the last digit of the extremes of the range of Integer
values and carefully avoids overflow. 该代码考虑了Integer
值范围的极值的最后一位,并小心地避免了溢出。
Integer.MAX_VALUE
is 2147483647
Integer.MAX_VALUE
是2147483647
That means: Integer.MAX_VALUE/10
is 214748364
这意味着: Integer.MAX_VALUE/10
为214748364
And you're about to do: rev = rev * 10 + pop
您将要做: rev = rev * 10 + pop
So if rev > 214748364
, then rev * 10
will cause overflow. 因此,如果rev > 214748364
,则rev * 10
将导致溢出。
Or if rev == 214748364
, then rev * 10
is 2147483640
, and rev * 10 + pop
will cause overflow if pop > 7
. 或者,如果rev == 214748364
,则rev * 10
为2147483640
,如果pop > 7
,则rev * 10
rev * 10 + pop
将导致溢出。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.