去掉字符串中的所有空格

       要去掉字符串中所有的空格，思路很自然：碰到一个空格，将其删掉，然后后面的字符前移，但当空字符比较多的时候，总这么移动，难道不复杂么？还是想想另外的办法吧。

      上面的方法时间复杂度比较高，下面，我们用空间来换时间：

#include <iostream>
using namespace std;

void deleteAllSpace(char str[])
{
int len = strlen(str);
char *p = new char[len + 1];
int i, j = 0;
for(i = 0; i < len; i++)
{
if(' ' != str[i])
{
p[j++] = str[i];
}
}

p[j] = '\0';
strcpy(str, p);
}

int main()
{
char str1[] = "abc";
char str2[] = "  abc";
char str3[] = "abc ";
char str4[] = " a b c ";
char str5[] = " a  b  c ";
char str6[] = " a  bcd  e ";

deleteAllSpace(str1);
deleteAllSpace(str2);
deleteAllSpace(str3);
deleteAllSpace(str4);
deleteAllSpace(str5);
deleteAllSpace(str6);

printf("%s\n", str1);
printf("%s\n", str2);
printf("%s\n", str3);
printf("%s\n", str4);
printf("%s\n", str5);
printf("%s\n", str6);

return 0;
}

     上面这个思路不错，但是，main中的程序太丑陋了，改为如下：

#include <iostream>
using namespace std;

void deleteAllSpace(char str[])
{
int len = strlen(str);
char *p = new char[len + 1];
int i, j = 0;
for(i = 0; i < len; i++)
{
if(' ' != str[i])
{
p[j++] = str[i];
}
}

p[j] = '\0';
strcpy(str, p);
delete []p;

}

int main()
{
char str[6][20] =
{
"abc",
"  abc",
"abc ",
" a b c ",
" a  b  c ",
" a  bcd  e "
};

int i;
for(i = 0; i < 6; i++)
{
deleteAllSpace(str[i]);
printf("%s\n", str[i]);

}

return 0;
}

       事实上，上述程序的空间复杂度较高，那有没有更好的办法呢？有，如下：

#include <iostream>
using namespace std;

void deleteAllSpace(char str[])
{
char *p = str;
int k = 0;

while('\0' != *p)
{
if(' ' != *p)
{
str[k++] = *p++;
}
else
{
p++;
}
}

str[k] = '\0';
}

int main()
{
char str[6][20] =
{
"abc",
"  abc",
"abc ",
" a b c ",
" a  b  c ",
" a  bcd  e "
};

int i;
for(i = 0; i < 6; i++)
{
deleteAllSpace(str[i]);
printf("%s\n", str[i]);

}

return 0;
}

        OK.