Say x & y are the integer variables.说x与y是整数的变数。 The challenge is to swap them without using a temporary variable.所面临的挑战是要交换他们不使用临时变量。

The solution is as simple as the problem itself:解决的办法是那么简单的问题本身：
x = x + y; x =用X + y ;
y = x - y;为Y = x -y ;
x = x - y; x = x -y ;

Update 1: Jack and Alexey pointed out a typo in my solution which has since been corrected.更新1 ：杰克和阿列克谢指出，错字在我的解决方案，其中已被纠正。 Please see their comments below.请参阅他们的意见如下。

The simplicity of the solution appeals to me.简单的解决方案呼吁我。 It clearly demonstrates the meaning of assignment operator ( “=” ).它清楚地表明所指的转让的营办商（ “ = ” ） 。

What if they are Strings?如果他们是字符串呢？
Note: You can use String methods in Java API.注意：您可以使用字符串的方法的Java API 。

Here is the solution:这里是解决方法：
x = x + y; x =用X + y ;
y = x.substring(0, x.indexOf(y));为Y = x.substring （ 0 ， x.indexof （元Y ） ） ;
x = x.substring(x.indexOf(y) + y.length()); x = x.substring （ x.indexof （元Y ） + y.length （ ） ） ;

Can you see the similarity?你能看到的相似性呢？

Can you provide a simpler solution to either of the above?你能提供一个更简单的解决办法上述任一？

Update 2: Robert just did (see his comment below).更新2 ：罗伯特只是没有（见他的评论下文） 。 He pointed out that the String solution doesn’t work when one string is contained in another.他指出，该字符串的解决办法行不通时，一个字符串是载于另一个。 His solution is:他的解决办法是：

x = x + y; x =用X + y ;
y = x.substring(0, x.length() - y.length());为Y = x.substring （ 0 ， x.length （ ） -y .length（ ） ） ;
x = x.substring(y.length()); x = x.substring （ y.length （ ） ） ;