[题目]去除字符串中多余空格

题目：将字符串中所有多于2个的空格变为1个空格

解答：C语言版本，我自己写的，多用了些内存，但是不用考虑那么多状态。：）

 

#include <stdlib.h>
#include <string.h>
#include <stdio.h>

char text[]="  I  love  Linux    hehe ";

int main(int argc, char **argv)
{
    int begin=0;
    int end = 0;
    char *p0 = (char *)malloc(sizeof(text)),*p1;
    int i,len=strlen(text);

    if(text[0] == ' ') begin=1;
    if(text[len-1] == ' ') end = 1;

    printf("Before converting...\n%s*\n",text);
    memset(p0,'\0',sizeof(p0));
    if(begin)
    {
        p0[0]='\0';
        strcat(p0," ");
    }
    
    for(i=0; i<len; i++)
    {
        if(text[i]==' ')
        {
            text[i]='\0';
        }else
        {
            p1 = &text[i];
            while(text[i] != ' ')
                i++;//越过所有非空格

            text[i]='\0';
            strcat(p0,p1);
            strcat(p0," ");
        }
    }
    if(!end)
        p0[strlen(p0)-1]='\0';

    printf("After converting...\n%s*\n",p0);
    free(p0);

    return 0;
}


Perl语言版本：

 

#!/usr/bin/perl

use strict;
use warnings;

my $str = " I love Linux hehe ";
print "Before converting...\n".$str."*\n";

$str =~ s/\ +/ /g;
print "After converting...\n".$str."*\n";