blocks|key|3065768|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3065769|+for(i+=0,+j=0;+i+<=+length+-+1;+i%2B%2B)
+{
++if(phrase[i]+!=+space)++++//Makes+the+phrase+without+spaces
++{
+++phrase2[j]+=+phrase[i];
+++j%2B%2B;
++}+
+}

+for(i+=+length+-1,+j+=+0;+i+>=+0;i--)
+{
++if(phrase[i]+!=+space)++++//Makes+the+phrase+backwards+an+without+spaces
++{
+++phrase3[j]+=+phrase[i];
+++j%2B%2B;
++}
+}

+length+=+j;|code-block|syntax|javascript|3065770|更新|offset|length|style|BOLD|3065771|作为对Praetorian的帖子的回应，这里是不复制字符串的代码。|3065772|#include+<stdio.h>
#include+<string.h>

int+main()
{
++int+i,+j,+length;
++char+space+=+'+';
++char+phrase[80];

++printf("Give+me+the+phrase:+");
++gets(phrase);
++length++++++=+strlen(phrase);

++for(+i+=+0,+j+=+length+-+1;+i+<+j;+i%2B%2B,+j--+)+{
++++while+(phrase[i]+==+space)+i%2B%2B;
++++while+(phrase[j]+==+space)+j--;
++++if(+phrase[i]+!=+phrase[j]+)+{
++++++printf("It's+not+a+palindrome\n");
++++++return+0;
++++}
++}

++printf("It's+a+palindrome\n");
++return+0;+
}|3065773|entityMap^0|0|0|0|2|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Try this:

<pre><code> for(i =0, j=0; i &lt;= length - 1; i++)
 {
 if(phrase[i] != space) //Makes the phrase without spaces
 {
 phrase2[j] = phrase[i];
 j++;
 } 
 }

 for(i = length -1, j = 0; i &gt;= 0;i--)
 {
 if(phrase[i] != space) //Makes the phrase backwards an without spaces
 {
 phrase3[j] = phrase[i];
 j++;
 }
 }

 length = j;
</code></pre>

Update

In response to Praetorian's post here's the code to do it without copying the string.

<pre><code>#include &lt;stdio.h&gt;
#include &lt;string.h&gt;

int main()
{
 int i, j, length;
 char space = ' ';
 char phrase[80];

 printf("Give me the phrase: ");
 gets(phrase);
 length = strlen(phrase);

 for( i = 0, j = length - 1; i &lt; j; i++, j-- ) {
 while (phrase[i] == space) i++;
 while (phrase[j] == space) j--;
 if( phrase[i] != phrase[j] ) {
 printf("It's not a palindrome\n");
 return 0;
 }
 }

 printf("It's a palindrome\n");
 return 0; 
}
</code></pre>

blocks|key|749687|text|在第二次循环之前，你想要设置j=0。在那之后它应该会工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|749688|PS:如果你通过打印出你的三个字符串来调试，你会在几分钟内弄清楚它。当您不知道哪里出了问题时，在中间步骤打印出变量的值，这样您就可以知道问题发生在哪里以及问题是什么。|749689|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|$]]

Before the 2nd loop you want to set j=0. It should work after that. 

PS: If you debugged by printing out your three strings, you would've figured it out in a matter of minutes. When you don't know what goes wrong, print out the values of variables at intermediate steps, so you know where your problem occurs and what it is.

blocks|key|3065828|text|你的问题已经被其他人回答了，但我发布这段代码是为了说明没有必要让phrase3副本保存颠倒的字符串。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3065829|#include+<stdio.h>
#include+<string.h>

int+main()
{

++int+i,+j,+length,+halfLength;
++char+space+=+'+';
++char+phrase1[80],+phrase2[80];

++printf("Give+me+the+phrase:+");
++gets(phrase1);
++length++++++=+strlen(phrase1);

++for(+i+=+0,+j+=+0;+i+<=+length;+%2B%2Bi+)+{
++++if(+phrase1[i]+!=+space+)+{++++//Makes+the+phrase1+without+spaces
++++++phrase2[j%2B%2B]+=+phrase1[i];
++++}
++}

++length++++++=+strlen(phrase2);
++halfLength++=+length+/+2;

++for(+i+=+0,+j+=+length+-+1;+i+<+halfLength;+%2B%2Bi,+--j+)+{
++++if(+phrase2[i]+!=+phrase2[j]+)+{
++++++printf("It's+not+a+palindrome\n");
++++++return+0;
++++}
++}

++printf("It's+a+palindrome\n");
++return+0;+
}|code-block|syntax|javascript|3065830|entityMap^0|W|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Your question has already been answered by others but I'm posting this code to show that it is not necessary to make the <code>phrase3</code> copy to hold the reversed string.

<pre><code>#include &lt;stdio.h&gt;
#include &lt;string.h&gt;

int main()
{

 int i, j, length, halfLength;
 char space = ' ';
 char phrase1[80], phrase2[80];

 printf("Give me the phrase: ");
 gets(phrase1);
 length = strlen(phrase1);

 for( i = 0, j = 0; i &lt;= length; ++i ) {
 if( phrase1[i] != space ) { //Makes the phrase1 without spaces
 phrase2[j++] = phrase1[i];
 }
 }

 length = strlen(phrase2);
 halfLength = length / 2;

 for( i = 0, j = length - 1; i &lt; halfLength; ++i, --j ) {
 if( phrase2[i] != phrase2[j] ) {
 printf("It's not a palindrome\n");
 return 0;
 }
 }

 printf("It's a palindrome\n");
 return 0; 
}
</code></pre>

blocks|key|976207|text|这是我想出来的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|976208|#include+<stdio.h>
void+main()+{
char+a[50],b[50];
int+i=0,j,ele,test=0,x;
while((a[i]=getchar())!='\n')+{
if(a[i]!='+'+&&+a[i]!=',')+//do+not+read+whitespaces+and+commas(for+palindromes+like+"Ah,+Satan+sees+Natasha")
i%2B%2B;
}
a[i]='\0';
ele=strlen(a);
//+Convert+string+to+lower+case+(like+reverse+of+Ava+is+avA+and+they're+not+equal)
for(i=0;+i<ele;+i%2B%2B)
if(a[i]>='A'&&a[i]<='Z')
a[i]+=+a[i]%2B('a'-'A');
x+=+ele-1;
for(j=0;+j<ele;+j%2B%2B)+{
b[j]+=+a[x];
x--;
}
for(i=0;+i<ele;+i%2B%2B)
if(a[i]==b[i])
test%2B%2B;
if(test==ele)
printf("You+entered+a+palindrome!");
else
printf("That's+not+a+palindrome!");
}|code-block|syntax|javascript|976209|可能不是回文的最好方法，但我很自豪我自己花了1个小时做这个:(+lol|976210|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is what I came up with: 

<pre><code>#include &lt;stdio.h&gt;
void main() {
char a[50],b[50];
int i=0,j,ele,test=0,x;
while((a[i]=getchar())!='\n') {
if(a[i]!=' ' &amp;&amp; a[i]!=',') //do not read whitespaces and commas(for palindromes like "Ah, Satan sees Natasha")
i++;
}
a[i]='\0';
ele=strlen(a);
// Convert string to lower case (like reverse of Ava is avA and they're not equal)
for(i=0; i&lt;ele; i++)
if(a[i]&gt;='A'&amp;&amp;a[i]&lt;='Z')
a[i] = a[i]+('a'-'A');
x = ele-1;
for(j=0; j&lt;ele; j++) {
b[j] = a[x];
x--;
}
for(i=0; i&lt;ele; i++)
if(a[i]==b[i])
test++;
if(test==ele)
printf("You entered a palindrome!");
else
printf("That's not a palindrome!");
}
</code></pre>

Probably not the best way for palindromes, but I'm proud I made this on my own took me 1 hour :( lol

blocks|key|3065848|text|为什么不使用std::stack呢？您将需要两个循环，每个循环迭代输入字符串的长度。在第一个循环中，遍历一次输入字符串，将每个字符压入堆栈。在第二个循环中，从堆栈中弹出一个字符，并将其与索引处的字符进行比较。如果你在循环结束之前得到一个不匹配的结果，你就不会有回文。这样做的好处是，您不必担心偶数/奇数长度的角点情况。它会正常工作的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3065849|(如果您愿意，可以使用一个堆栈(LIFO)和一个队列(FIFO)，但这并不会实质上改变算法)。|3065850|下面是实现：|3065851|bool+palindrome(const+char+*s)
{
++++std::stack<char>+p;+//+be+sure+to+#include+<stack>

++++for(int+i+=+0;+s[i]+!=+0;+i%2B%2B)
++++++++p.push(s[i]);

++++for(int+i+=+0;+s[i]+!=+0;+i%2B%2B)
++++{
++++++++if(p.top()+!=+s[i])
++++++++++++return+false;+//+not+a+palindrome!

++++++++p.pop();
++++}++++

++++return+true;
}|code-block|syntax|javascript|3065852|跳过空格作为练习留给读者;)|3065853|entityMap^0|6|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|Y|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Why not use a <code>std::stack</code>? You will need two loops, each iterating the length of the input string. In the first loop, go through the input string once, pushing each character ont the stack. In the second loop, pop a character off the stack and compare it with the character at the index. If you get a mismatch before the loop ends, you don't have a palindrome. The nice thing with this is that you don't have to worry about the even/odd length corner-case. It will just work.

(If you are so inclined, you can use one stack (LIFO) and one queue (FIFO) but that doesn't substantially change the algorithm).

Here's the implementation:

<pre><code>bool palindrome(const char *s)
{
 std::stack&lt;char&gt; p; // be sure to #include &lt;stack&gt;

 for(int i = 0; s[i] != 0; i++)
 p.push(s[i]);

 for(int i = 0; s[i] != 0; i++)
 {
 if(p.top() != s[i])
 return false; // not a palindrome!

 p.pop();
 } 

 return true;
}
</code></pre>

Skipping spaces is left as an exercise to the reader ;)

I'm trying to determine if a phrase is a palindrome (a word that is the same from left to rigth) or not but i can't make it work. What's wrong?, i can't use pointers or recursion or string type variables

<pre><code>#include &lt;stdio.h&gt;

#include &lt;string.h&gt;

int main()

{

 int i,j = 0,length;
 char space = ' ';
 char phrase [80],phrase2[80],phrase3[80];

 printf("Give me the phrase: ");
 gets(phrase);
 length = strlen(phrase);

 for(i =0; i &lt;= length - 1; i++)
 {
 if(phrase[i] != space) //Makes the phrase without spaces
 {
 phrase2[i] = phrase[i];
 j++;
 }
 }

 for(i = length -1; i &gt;= 0;i--)
 {
 if(phrase[i] != space) //Makes the phrase backwards an without spaces
 {
 phrase3[j] = phrase[i];
 j++;
 }
 }

 length = strlen(phrase2);

 for(i =0; i &lt;= length -1;i++) //Compare the phrases to know if they are the same
 {
 if(phrase2[i] != phrase3[i])
 {
 printf("It's not a palindrome\n"); 
 return 0;
 }
 }
 printf("It's a palindrome\n");
 return 0; 
}
</code></pre>

Palindrome in C without pointers and recursion

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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我正在尝试确定一个短语是否是回文(从左到右都一样的单词)，但是我不能让它工作。怎么了?我不能使用指针、递归或字符串类型的变量#include <stdio.h>#include <string.h>int main(){ int i,j = 0,length; char space = ' '; char phrase...

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