blocks|key|4840214|text|我最初的简单解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4840215|1.+print(+(abs(Number)%2BNumber)+/+2+)|code-block|syntax|javascript|4840216|这种解决方案在大多数情况下都有效，除非数字非常大(超过最大数字的一半，例如>=+MAX_INT/2)，在这种情况下，相加可能会导致溢出。|4840217|以下解决方案解决了溢出问题：|4840218|2.+print(+(abs(Number)/2)+%2B+(Number/2)+)|4840219|但是，可能有这样一种情况，数字是并且必须保持整数，并且除法运算符(/)是整数除法，因此7/2=3。在这种情况下，解决方案2将不起作用，因为如果为Number=7，它将打印6(对于这种情况，解决方案1将工作得很好)。|4840220|因此，如果我们需要同时处理大数和整数算术，下面的怪胎就来了，在奇数整数的情况下，添加了在除以2的过程中可能丢失的1的补偿：|4840221|3.+print(+
++++(+(abs(Number)/2)%2B(Number/2)+)+%2B
++++((
++++++++(Number-(2*(Number/2)))+%2B+
++++++++(abs(Number)-(2*(abs(Number)/2)))
++++)+/+2)
++++)++++++++|4840222|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

My original simple solution:

<pre><code>1. print( (abs(Number)+Number) / 2 )
</code></pre>

That solution would work in most cases, unless Number is very large (more than half the maximum e.g. Number >= MAX_INT/2) in which case the addition may cause overflow.

The following solution solves the overflow problem:

<pre><code>2. print( (abs(Number)/2) + (Number/2) )
</code></pre>

However, there may be a case in which Number is and must remain integer, and the division operator (/) is integer division, so that 7/2=3. In this case solution 2 won't work because if Number=7 it will print 6 (for this case solution 1 will work just fine). 

So if we need to deal with both large numbers AND integer arithmetic, the following monstrosity comes to the rescue, adding compensation for the 1 that may be lost in the division by 2 in case of odd integer:

<pre><code>3. print( 
 ( (abs(Number)/2)+(Number/2) ) +
 ((
 (Number-(2*(Number/2))) + 
 (abs(Number)-(2*(abs(Number)/2)))
 ) / 2)
 ) 
</code></pre>

blocks|key|4840154|text|print+max(0,+number)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4840155|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>print max(0, number)
</code></pre>

blocks|key|1254257|text|假设该数字由一个8位二进制补码整数表示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1254258|包括0在内的正数都将MSB设置为0。|1254259|所有负数都将MSB设置为1。|1254260|因此，我们采用MSB的补码，将其扩展到完整的8位，并将其与原始数字进行逐位与运算，例如|1254261|正面：|1254262|00110101+->最高有效位为0|1254263|11111111扩展的->补丁|1254264|00110101+->按位AND或以上|1254265|否定：|1254266|10110101+->+MSB为1|1254267|00000000扩展的->补丁|1254268|00000000+->按位AND或以上|1254269|不需要比较-我假设这是按位的，并不是严格意义上的比较。|1254270|另外，很抱歉没有代码，但是你明白我的意思。|1254271|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|15|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|16|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|17|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|18|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|19|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1A|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1B|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1C|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1D|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1E|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1G|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|12|$]]

Let's say that number is represented by an 8-bit two's complement integer.

Positive numbers including 0 all have the MSB set to 0.

Negative numbers all have the MSB set to 1.

So we take the complement of the MSB, extend it to the full 8 bits, and bitwise AND it with the original number, e.g.

Positive: 

00110101 -> MSB is 0

11111111 -> complement of MSB extended

00110101 -> bitwise AND of above

Negative:

10110101 -> MSB is 1

00000000 -> complement of MSB extended

00000000 -> bitwise AND of above

No comparisons needed - I'm kind of assuming that bitwise AND isn't strictly a comparison.

Also, sorry for the lack of code, but you get the idea.

blocks|key|776158|text|我还没有看到一个对整个域有效的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|776159|另一种解决方案是调用一个函数，该函数在输入值为0或小于0时引发异常。然后捕获异常并打印0。|776160|也可以使用函数符号，如果输入<0，则返回-1，如果输入为0，则返回0，否则返回1。|776161|print+((sign(x)%2B1)+*+sign(x)+/+2)+*+x.|code-block|syntax|javascript|776162|sign可以是-1、0或1，因此((+sign(x)+%2B1)+*sign(X)/+2)可以具有下列值：|776163|-1+->+((-1%2B1)*-1)/2+=+0
+0+->+((0%2B1)*0)/2+=+0
+1+->+((1%2B1)+*+1)/2++=+1|776164|另一种方法是创建一个查找表，将所有非负数映射到它们自己，其余的映射到0。|776165|但是，在我看来，最初的功能要清晰得多。那么为什么要违反KISS原则呢。|776166|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I haven't seen a solution yet that is valid for the complete domain.

An other solution is to call a function that raises an exception if the input value is 0 or below 0. Then catch the exception and print 0.

Or you can use a function Sign, that returns -1 if the input is &lt;0, 0 if it's 0 and 1 otherwise.

<pre><code>print ((sign(x)+1) * sign(x) / 2) * x.
</code></pre>

sign can be -1, 0 or 1, so ((sign(x)+1) * sign(x) / 2) can have the following values:

<pre><code>-1 -&gt; ((-1+1)*-1)/2 = 0
 0 -&gt; ((0+1)*0)/2 = 0
 1 -&gt; ((1+1) * 1)/2 = 1
</code></pre>

Another method is to create a lookup table that maps all non negative numbers to themself and the rest to 0.

But, in my opinion, the original function is much clearer. So why violate the KISS principle.

blocks|key|1248752|text|类似于公认的答案。虽然绝对值是通过比较实现的，但也更容易溢出：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248753|print(+(sqrt(Number+*+Number)+%2B+x)+/+2)；|1248754|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Similar to the accepted answer. Although acceptable where absolute value is implemented using comparisons, but also more prone to overflow:

print( (sqrt(Number * Number) + x) / 2);

blocks|key|3994748|text|假设C或C%2B%2B：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3994749|switch+((unsigned+long)Number+&+~(unsigned+long)LONG_MAX)+{
++++case+0:
++++++++printf("%25d\n",+Number);
++++++++break;
++++default:
++++++++printf("0\n",+Number);
++++++++break;
}|code-block|syntax|javascript|3994750|如果您认为switch是一个条件运算符，那么可以尝试这样做：|3994751|unsigned+long+flag+=+(unsigned+long)Number+&+~(unsigned+long)LONG_MAX;
flag+/=+(unsigned+long)LONG_MAX+%2B+1;
flag+=+1+-+flag;
printf("%25d\n",+Number+*+flag);|3994752|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Assuming C or C++:

<pre><code>switch ((unsigned long)Number &amp; ~(unsigned long)LONG_MAX) {
 case 0:
 printf("%d\n", Number);
 break;
 default:
 printf("0\n", Number);
 break;
}
</code></pre>

If you consider switch to be a conditional operator, then try this:

<pre><code>unsigned long flag = (unsigned long)Number &amp; ~(unsigned long)LONG_MAX;
flag /= (unsigned long)LONG_MAX + 1;
flag = 1 - flag;
printf("%d\n", Number * flag);
</code></pre>

Do the below without any conditional or comparison operator.

<pre><code>if (Number &lt;= 0)
{
 Print '0';
}
else
{
 print Number;
}
</code></pre>

thanks..

Interview qns...Do the below without any conditional or comparison operator

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在不使用任何条件或比较运算符的情况下执行以下操作。if (Number <= 0){ Print '0';}else{ print Number;}谢谢..

问Interview qns...Do下面不带任何条件或比较运算符
EN

回答 6

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问Interview qns...Do下面不带任何条件或比较运算符EN

回答 6

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问Interview qns...Do下面不带任何条件或比较运算符
EN