小编cod*_*ict的帖子

期望以下例程在单个链表的开头删除节点,但有时会失败.

void remove_from_front(node *start)
{
    delete start;
    start = start->link;
    print_list(start);
}

 127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/ HTTP/1.1" 200 169 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/.treeinfo HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Fedora HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Server HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Client HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/RedHat HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/CentOS HTTP/1.1" …

我有以下问题
给定一个字符串,返回一个"清理"字符串,其中相同的字符相同的字符已减少为一个字符.所以"yyzzza"产量"yza".

stringClean("yyzzza") ? "yza"      
stringClean("abbbcdd") ? "abcd"       
stringClean("Hello") ? "Helo"

我正在尝试输入代码 stringClean("abbbcdd") ? "abcd"

我的代码在下面.我在进行相邻字符比较后得到部分附加字符串,因此截至现在我得到附加的stringBuilder "sb=abc",这不是正确的输出我应该得到输出"abcd",

class cleanString{

    public static String stringClean(String str){
        int startIndex = str.indexOf(str);
        char startChar = '\u0000';
        char adjacentChar = '\u0000';
        System.out.println("startIndex-->" + startIndex);
        final StringBuilder sb = new StringBuilder();

        for(startIndex = 0; startIndex < str.length(); startIndex += 1){
            startChar = str.charAt(startIndex);
            System.out.println("startIndex ::" + startIndex);
            System.out.println("startChar ::" + startChar);

            final int adjacentPosition = startIndex + 1;
            System.out.println("adjacentPosition …

如果我有一个Java类,如下所示,相同的Perl表示是什么？

class Temp {
  int my_int;
  int my_array[];

  Temp() {
  }
};

具体来说,我有兴趣知道如何使用数组作为对象变量？

我是指针和链表的新手,但我正在尝试编写一个简单的程序,将文本文件中的数据读入链表.我输入功能有问题.它看起来像这样:

DVDNode* CreateList(string fileName)
{
    ifstream inFile;
    inFile.open(fileName.c_str());

    DVDNode* head = NULL;
    DVDNode* dvdPtr;
    dvdPtr = new DVDNode;

    while(inFile && dvdPtr != NULL)
    {
        getline(inFile, dvdPtr -> title);
        getline(inFile, dvdPtr -> leadActor);
        getline(inFile, dvdPtr -> supportingActor);
        getline(inFile, dvdPtr -> genre);
        cin >> dvdPtr -> year;
        cin >> dvdPtr -> rating;
        getline(inFile, dvdPtr -> synopsis);

        dvdPtr -> next = head;
        head = dvdPtr;
        dvdPtr = new DVDNode;
    }

    delete dvdPtr;
    inFile.close();

    return head;
}

我也有一个输出函数,如下所示:

void OutputList(DVDNode* head, string outputFile)
{ …

我尝试了一些事情,但我收到了错误

typedef '\0' DEFAULT_VALUE;   

错误:'\ 0'之前预期的未命名id

typedef NULL DEFAULT_VALUE;   

错误:__null之前预期的未命名id

我在这做错了什么？

我有一个如下所示的序列:

my $seq = "D\IKLR\LK/Q";

我想要做的是将序列分解为单个字母.

所以我希望得到:

my $var = ['D', '\', 'I', 'K', 'L', 'R', '\', 'L', 'K','/' ,'Q'];

但为什么这样做不会:

 my @chars = split(//,$seq);
 print Dumper \@chars;

它改为:

my $var = ['D', '\\', 'I', 'K', 'L', 'R', '\\', 'L', 'K', 'Q'];

做正确的方法是什么？

似乎无法解决为什么我在打印内容时从此链接列表结构获取垃圾输出.

我的目标是在列表中添加任何内容,一些字符串,char by char,它应该反向打印出来.我使用额外的头部+尾部结构的原因是我可以打印出反向输入的订单行.

typedef struct List {
 char c;
 struct List *next;
}List;

typedef struct  {
 List *head;
 List *tail;
}FullList;

List* InsertList(int hd, List* t1) {
 List *t = (List*)calloc(1,sizeof(List));
 t->c = hd; 
 t->next = t1;
return t;
}


FullList addToStart(FullList c, char element) {
 if (c.head == NULL) {
    c.head = c.tail = InsertList(element, NULL);
 }else {
    c.head = InsertList(element, c.head);
 }
return c;
}

int main(void) {
 FullList InOrder;
 FullList Reverse;
 InOrder.head = NULL;
 Reverse.head = …

Array ( [0] => Array ( [OPT] => 65 ) )

如何65在PHP中获取变量

请帮帮我....

这是我关于 setter 和 getter 的任务，由于某种原因它不起作用。谁能检查一下我的问题是什么？谢谢你。

public class FlightTest
{
    public static void main (String [] args)
    {

        String name;
        String number;
        String Orig;
        String Desti;

        Scanner scan = new Scanner (System.in);

        Flight data = new Flight ();

        System.out.print ("Airline Name: ");
        String AirlineName = scan.nextLine ();
        data.setAirlineName (name);

        System.out.print ("Flight Number: ");
        String FlightNumber = scan.nextLine ();
        data.setFlightNumber (number);

        System.out.print ("Origin: ");
        String Origin = scan.nextLine();
        data.setOrigin (Orig);

        System.out.print ("Destination: ");
        String Destination = scan.nextLine ();
        data.setDestination (Desti);

        System.out.println (data); …