有人可以用一段代码帮助我吗？ [英] Could someone help me with a fragment of code?

问题描述

全部晚上，



对于作业，我必须编写一个例程，删除同一个字符的所有第二次出现。例如，如果我输入apple，输出将是aple，其中只有字母表中的每个字母中的一个在输出内。

另一个例子：aaabbb给出ab。





我写了这段代码，我不明白为什么它不起作用。那些对c有更多了解的人可以帮个忙吗？谢谢。

Evening all,

For an assignment I have to write a routine that will delete all the second occurrence of the same character. For example if i input apple the output will be aple where only one of each letter from the alphabet is within the output.
another example: aaabbb gives ab.


I have written this code which I can't see why it isnt working. Could someone with a bigger knowledge of c give me a hand? Thank you.

void dublicateRemoved( char key[] )
	{
		int letterFoundSame = 0;
		for(int i = 0; i < 16; i++) //Go through each letter key // 0 to 15
   		{           	
      		for(int x = 0; x <= i; x++) //Go through all letters so far.
         		{
            		if(key[i] == cutKey[x]) //True at anypoint then we disgard it as we want no dublicates.
               		letterFoundSame = 1;
                  else
                  	letterFoundSame = 0;
            	}
       		//So if a letter has been found to be the same we dont want to add the letter to the list     
       		if(letterFoundSame == 0) //Not been repeated
         		cutKey[i] = key[i]; 
      	}
}

推荐答案

如果N是最大长度，请制作两个char数组，比如A [N]和B [N]。

在迭代A时，使用辅助函数检查char c是否在B中：

如果不将c添加到B并增加索引。继续直到c = A [i]为空（== 0）

if N is max length, make two char arrays, say A[N] and B[N].
while iterating through A, make helper function to check if char c is in B:
if not add c to B and increment index. keep going until c=A[i] is null (==0)

你的cutKey数组需要它自己的索引 - 所以你有

Your cutKey array needs its own index - so you have

if(letterFoundSame == 0) //Not been repeated
                cutKey[cutKeyIndex++] = key[i];

那么你的'通过所有字母到目前为止'循环使用cutKeyIndex（不是我）作为其限制。



你在测试中不需要其他 - 只需在内循环之前设置为0然后

then your 'go through all the letters so far' loop uses the cutKeyIndex (not i) as its limit.

You don't need an else in the test - just set to 0 before the inner loop then

if(key[i] == cutKey[x]) //True at anypoint then we disgard it as we want no dublicates.
{
                    letterFoundSame = 1;
                    break;
}

否则你找到一个匹配，将标志设置为1，然后继续查看下一个字符，找不到匹配，并将其设置回0。

otherwise you find a match, set the flag to 1, then keep looking at the next character, find no match, and set it back to 0.

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