entityMap|blocks|key|ch5u2|text|这是非常直接的。添加它的主要目的是为了不会遇到整数内存越界异常。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|14tn7|Line+1:+++rev+==+Integer.MAX_VALUE+/+10
Line+2:++rev+=+rev+*+10+%2B+pop;|code-block|syntax|javascript|5h6lp|让我们假设上面的情况对于某些值是真的。Line+1+means+that+10*rev+=+Integer.MAX_VALUE.第2行:+rev是一个整数。如果按照第1行，我们遇到了rev+=+Integer.MAX_VALUE,的情况，如果我们在它上面添加一个pop，我们将遇到整数内存越界异常。为了避免这种情况，添加了第1行。确保执行第1行的示例测试用例是"Pass+Integer.MAX_VALUE"+as+an+input+to+reverse()+function.|offset|length|style|CODE|fk3n1^0|0|0|J|19|2J|O|50|1N|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|Q|9|@$J|R|K|S|L|M]|$J|T|K|U|L|M]|$J|V|K|W|L|M]]|A|@]|B|$]]|$2|N|4|-4|6|7|8|X|9|@]|A|@]|B|$]]]]

It's pretty straight forward. Its mainly added so that you don't hit integer memory out of bound exception. 

<pre><code>Line 1: rev == Integer.MAX_VALUE / 10
Line 2: rev = rev * 10 + pop;
</code></pre>

Let's assume the above case is true for some value. <code>Line 1 means that 10*rev = Integer.MAX_VALUE.</code> Line 2: rev is an integer. if as per line 1, we have encountered a case where <code>rev = Integer.MAX_VALUE,</code> if we add a pop on top of it, we would encounter integer memory out of bound exception. To avoid that, Line 1 is added. Sample test case to make sure Line 1 is executed is <code>"Pass Integer.MAX_VALUE" as an input to reverse() function.</code>

entityMap|blocks|key|h171|text|reverse(int+x)函数用于反转x的十进制数字。例如，如果为x+=+102，则返回201；如果为x+=+-102，则返回-201。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3tpq6|当计算出x的倒数时，它存储在rev中。例如，对于x+=+102，rev采用连续的值2、20和201。|8arth|在每次迭代中，rev与10相乘，并向其添加一个数字。当然，rev不能大于Integer.MAX_VALUE+(2147483647)。因此，在将rev与10相乘之前，我们检查将其与10相乘并添加pop是否会使其大于Integer.MAX_VALUE。|4217p|我们首先检查rev是否大于Integer.MAX_VALUE+/+10+(214748364)。如果它大于，则反向整数不存在。如果不是，则rev小于或等于Integer.MAX_VALUE+/+10。如果它小于Integer.MAX_VALUE+/+10，那么即使我们将它乘以10，我们也可以向它添加任何数字(pop)，并且我们不会超过Integer.MAX_VALUE。但是，如果它等于Integer.MAX_VALUE+/+10，那么我们必须确保pop不是>+7+(2147483647+-+214748364+*+10)，因为否则我们将超过Integer.MAX_VALUE。|8f3g6|同样的道理也适用于Integer.MIN_VALUE+(2147483648)。|d7mj6^0|0|E|K|1|Y|7|19|3|1G|8|1S|4|0|4|1|E|3|O|7|W|3|15|1|17|2|1A|3|0|7|3|B|2|T|3|10|H|1J|A|20|3|24|2|2I|2|2P|3|2Z|H|0|6|3|D|M|11|9|1X|3|25|M|2X|M|3U|2|4B|3|4O|H|5E|M|69|3|6E|3|6J|R|7L|H|0|9|H|S|A|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@$A|Q|B|R|C|D]|$A|S|B|T|C|D]|$A|U|B|V|C|D]|$A|W|B|X|C|D]|$A|Y|B|Z|C|D]|$A|10|B|11|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|12|9|@$A|13|B|14|C|D]|$A|15|B|16|C|D]|$A|17|B|18|C|D]|$A|19|B|1A|C|D]|$A|1B|B|1C|C|D]|$A|1D|B|1E|C|D]|$A|1F|B|1G|C|D]]|E|@]|F|$]]|$2|I|4|J|6|7|8|1H|9|@$A|1I|B|1J|C|D]|$A|1K|B|1L|C|D]|$A|1M|B|1N|C|D]|$A|1O|B|1P|C|D]|$A|1Q|B|1R|C|D]|$A|1S|B|1T|C|D]|$A|1U|B|1V|C|D]|$A|1W|B|1X|C|D]|$A|1Y|B|1Z|C|D]|$A|20|B|21|C|D]]|E|@]|F|$]]|$2|K|4|L|6|7|8|22|9|@$A|23|B|24|C|D]|$A|25|B|26|C|D]|$A|27|B|28|C|D]|$A|29|B|2A|C|D]|$A|2B|B|2C|C|D]|$A|2D|B|2E|C|D]|$A|2F|B|2G|C|D]|$A|2H|B|2I|C|D]|$A|2J|B|2K|C|D]|$A|2L|B|2M|C|D]|$A|2N|B|2O|C|D]|$A|2P|B|2Q|C|D]|$A|2R|B|2S|C|D]|$A|2T|B|2U|C|D]]|E|@]|F|$]]|$2|M|4|N|6|7|8|2V|9|@$A|2W|B|2X|C|D]|$A|2Y|B|2Z|C|D]]|E|@]|F|$]]|$2|O|4|-4|6|7|8|30|9|@]|E|@]|F|$]]]]

The <code>reverse(int x)</code> function reverses the decimal digits of <code>x</code>. For example, if <code>x = 102</code>, it returns <code>201</code>, and if <code>x = -102</code>, it returns <code>-201</code>.

When the reverse of <code>x</code> is computed, it is stored in <code>rev</code>. For example, for <code>x = 102</code>, <code>rev</code> takes the successive values <code>2</code>, <code>20</code>, <code>201</code>.

At each iteration, <code>rev</code> is multiplied by <code>10</code>, and one digit is added to it. Of course, <code>rev</code> cannot be greater than <code>Integer.MAX_VALUE</code> (<code>2147483647</code>). So, before multiplying <code>rev</code> by <code>10</code>, we check whether multiplying it by <code>10</code> and adding <code>pop</code> would make it greater than <code>Integer.MAX_VALUE</code>.

We first check whether <code>rev</code> is greater than <code>Integer.MAX_VALUE / 10</code> (<code>214748364</code>). If it is greater, the reverse integer does not exist. If not, then <code>rev</code> is less than or equal to <code>Integer.MAX_VALUE / 10</code>. If it is less than <code>Integer.MAX_VALUE / 10</code>, then even if we multiply it by <code>10</code>, we can add any digit (<code>pop</code>) to it, and we will not exceed <code>Integer.MAX_VALUE</code>. However, if it is equal to <code>Integer.MAX_VALUE / 10</code>, then we must that make sure that <code>pop</code> is not <code>&gt; 7</code> (<code>2147483647 - 214748364 * 10</code>) because we would otherwise exceed <code>Integer.MAX_VALUE</code>.

The same reasoning applies to <code>Integer.MIN_VALUE</code> (<code>2147483648</code>).

This is code from Leetcode solution. So I don't understand the if condition, why Integer.MAX_VALUE has to be divided by 10 (Interger.MAX_VALUE / 10) ?

Thank You!

<pre><code>class Solution {
 public int reverse(int x) {
 int rev = 0;
 while (x != 0) {
 int pop = x % 10;
 x /= 10;
 if (rev &gt; Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 &amp;&amp; pop &gt; 7)) return 0;
 if (rev &lt; Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 &amp;&amp; pop &lt; -8)) return 0;
 rev = rev * 10 + pop;
 }
 return rev;
 }
}


</code></pre>

Java Integer Overflow and why Integer.MAX_VALUE / 10?

 这是来自Leetcode解决方案的代码。所以我不理解if条件，为什么Integer.MAX_VALUE必须被10整除(Interger.MAX_VALUE / 10)？ 谢谢! class Solution {    public int reverse(int x) {        int rev = 0;   ...

问Java溢出，为什么是Integer.MAX_VALUE / 10？
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