entityMap|blocks|key|16v6f|text|很简单，因为当x==2和y==1条件x>y为true时，"("+%2B+(x-1)+%2B+","+%2B+y+%2B+")+"返回(1,1)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|fl4lu^0|7|4|C|4|I|3|S|S|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@$A|I|B|J|C|D]|$A|K|B|L|C|D]|$A|M|B|N|C|D]|$A|O|B|P|C|D]]|E|@]|F|$]]|$2|G|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

Simply because when <code>x==2</code> and <code>y==1</code> condition <code>x&gt;y</code> is true and <code>&quot;(&quot; + (x-1) + &quot;,&quot; + y + &quot;) &quot;</code> returns (1,1)

entityMap|blocks|key|7dq24|text|您的定义应如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4mv5d|private+static+String+createPath(int+x,+int+y,+String+path)+{
++++if+(x+<=+y)+{
++++++++return+path;
++++}
++++path+%2B=+"("+%2B+(x+-+1)+%2B+","+%2B+y+%2B+")+";
++++return+createPath(x+-+1,+y,+path);
}|code-block|syntax|javascript|4o92k|这个定义将确保只要返回x+<=+y，就会返回path的值。|offset|length|style|CODE|anc6k^0|0|0|B|6|M|4|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]|$J|T|K|U|L|M]]|A|@]|B|$]]|$2|N|4|-4|6|7|8|V|9|@]|A|@]|B|$]]]]

Your definition should be as follows:
<pre><code>private static String createPath(int x, int y, String path) {
 if (x &lt;= y) {
 return path;
 }
 path += &quot;(&quot; + (x - 1) + &quot;,&quot; + y + &quot;) &quot;;
 return createPath(x - 1, y, path);
}
</code></pre>
This definition will ensure that as soon as <code>x &lt;= y</code>, the value of <code>path</code> will be returned.

entityMap|blocks|key|e6teg|text|正确的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|67h7n|++++private+static+String+createPath(int+x,+int+y,+String+path)+{
++++++++if+(x+>+(y+%2B+1))
++++++++++++return+createPath(x+-+1,+y,+path+%2B+"("+%2B+(x+-+1)+%2B+","+%2B+y+%2B+")+");
++++++++else+
++++++++++++return+path+%2B+"("+%2B+(x+-+1)+%2B+","+%2B+y+%2B+")+";
++++}|code-block|syntax|javascript|9idoh|注意，else分支中的if+(x+>+(y+%2B+1))和(x+-+1)。|offset|length|style|CODE|3vnue|函数的执行顺序如下：|9qr2a|createPath(5,+1)+=>+createPath(4,+1,+"(4,1)");
createPath(4,+1)+=>+createPath(3,+1,+"(3,1)");
createPath(3,+1)+=>+createPath(2,+1,+"(2,1)");
createPath(2,+1)+=>+createPath(1,+1,+"(1,1)");
//+and+the+last+one+in+the+else+branch
createPath(1,+1)+=>+createPath(1,+1,+"(1,1)");|avi0g|正如您所看到的，您需要在else分支中添加(x+-+1)，并修改if语句，以便更早地到达else分支，以防止它低于1。|3dkj3^0|0|0|B|G|S|7|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|U|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|V|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|W|9|@$J|X|K|Y|L|M]|$J|Z|K|10|L|M]]|A|@]|B|$]]|$2|N|4|O|6|7|8|11|9|@]|A|@]|B|$]]|$2|P|4|Q|6|E|8|12|9|@]|A|@]|B|$F|G]]|$2|R|4|S|6|7|8|13|9|@]|A|@]|B|$]]|$2|T|4|-4|6|7|8|14|9|@]|A|@]|B|$]]]]

Correct solution:
<pre><code> private static String createPath(int x, int y, String path) {
 if (x &gt; (y + 1))
 return createPath(x - 1, y, path + &quot;(&quot; + (x - 1) + &quot;,&quot; + y + &quot;) &quot;);
 else 
 return path + &quot;(&quot; + (x - 1) + &quot;,&quot; + y + &quot;) &quot;;
 }

</code></pre>
note, <code>if (x &gt; (y + 1))</code> and <code>(x - 1)</code> in else branch.
the execution order of your function is the following:
<pre><code>createPath(5, 1) =&gt; createPath(4, 1, &quot;(4,1)&quot;);
createPath(4, 1) =&gt; createPath(3, 1, &quot;(3,1)&quot;);
createPath(3, 1) =&gt; createPath(2, 1, &quot;(2,1)&quot;);
createPath(2, 1) =&gt; createPath(1, 1, &quot;(1,1)&quot;);
// and the last one in the else branch
createPath(1, 1) =&gt; createPath(1, 1, &quot;(1,1)&quot;);
</code></pre>
as you can see you need to add (x - 1) in the else branch and modify your if statement in such way that you get to the else branch earlier to prevent it from going to below 1.

entityMap|blocks|key|eugc1|text|事实证明，打印两次最终值的原因是因为else返回语句不必要地添加了(x，y)的最终值。例如，|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2k3rt|else+
++++++return+path+%2B=+"("+%2B+x+%2B+","+%2B+y+%2B+")+";|code-block|syntax|javascript|1vbku|一旦x不再大于y，就会执行上面的代码，这意味着通过createPath(int+x,+int+y,+String+path)传递的路径的值包含彼此相等的(x，y)。然而，上面的else语句所做的是再次将x和y的值添加到path，而所有需要做的就是return+path;。|2bdk0^0|I|4|0|0|2|1|7|1|P|11|2T|1|2V|1|31|4|3F|C|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@$A|P|B|Q|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|R|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|S|9|@$A|T|B|U|C|D]|$A|V|B|W|C|D]|$A|X|B|Y|C|D]|$A|Z|B|10|C|D]|$A|11|B|12|C|D]|$A|13|B|14|C|D]|$A|15|B|16|C|D]]|E|@]|F|$]]|$2|N|4|-4|6|7|8|17|9|@]|E|@]|F|$]]]]

Turns out that the reason why the final value was being printed twice was because the <code>else</code> return statement was adding the final value of (x,y) unnecessarily. For example,
<pre><code>else 
 return path += &quot;(&quot; + x + &quot;,&quot; + y + &quot;) &quot;;
</code></pre>
The above code is executed once <code>x</code> is no longer greater than <code>y</code> , meaning the value of path that was passed through <code>createPath(int x, int y, String path)</code> contains (x,y) which are equal to one another. However, what the else statement above does is once again add the values of <code>x</code> and <code>y</code> to <code>path</code>, when all that needs to be done is <code>return path;</code>.

The following code recursively reduces x by 1 until it is equal to y. It prints a string representing the recursion process.
I do not want the starting value of x to be printed in the string.
<pre><code>private static String createPath(int x, int y, String path) {
 if (x &gt; y) {
 path += &quot;(&quot; + (x-1) + &quot;,&quot; + y + &quot;) &quot;;
 return createPath(x - 1, y, path);
 }
 else 
 return path += &quot;(&quot; + x + &quot;,&quot; + y + &quot;) &quot;;
 }
</code></pre>
When I input something such as:
<pre><code>System.out.println(createPath(5, 1, &quot;&quot;));
</code></pre>
I get the following output:
<pre><code>(4,1) (3,1) (2,1) (1,1) (1,1) 
</code></pre>
Why is the final value (1,1) being printed twice? I noticed that when I remove (x-1) from
<pre><code>path += &quot;(&quot; + (x-1) + &quot;,&quot; + y + &quot;) &quot;;
</code></pre>
The output becomes:
<pre><code>(5,1) (4,1) (3,1) (2,1) (1,1)
</code></pre>
But once again, I do not want (5,1) to be printed in the string. Also, I have had a look at <a href="https://stackoverflow.com/questions/57029711/why-is-my-recursive-loop-printing-the-last-value-twice-at-the-end">Why is my recursive loop printing the last value twice at the end?</a> but it did not seem to help me.

Why is my recursive loop printing the same value twice? - JAVA

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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 下面的代码递归地将x减去1，直到它等于y。它打印一个表示递归过程的字符串。我不希望在字符串中打印x的起始值。 private static String createPath(int x, int y, String path) {        if (x > y) {            path += "("...

问为什么我的递归循环两次打印相同的值？
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问为什么我的递归循环两次打印相同的值？EN