对于学校项目,我正在制作RPN计算器。这个计算器有一个抽象类Stack,其中三个不同的堆栈实现派生自。其中一个是链表列表。对于这些堆栈,我必须显示堆栈中的值。为此,我使用一个数组。对于我的arraystack和列表堆栈,这很简单,但我无法弄清楚如何将链表转换为数组。最好的方法是什么?
这是我用于链接列表的代码。
langs = {
"CRA": {flag: '',cs:'4262',as:''},
"Reviews": {flag: '',cs:'72074301',as:''},
"Europe": {
"Bulgarian": {flag: 'bg',cs:'4668',as:''},
"Croatian": {flag: 'hr',cs:'4671',as:'4707'},
"Czech": {flag: 'cz',cs:'4365',as:'4553'},
"Danish": {flag: 'dk',cs:'4357',as:'4526'},
"Dutch": {flag: 'nl',cs:'4343',as:'4511'},
"English": {flag: 'gb',cs:'4164',as:'4482'},
"Finnish": {flag: 'fi',cs:'4092',as:''},
"Flemish": {flag: 'be',cs:'4500',as:''},
"French": {flag: 'fr',cs:'4344',as:'4781'},
"German": {flag: 'de',cs:'4345',as:'4473'},
"Hungarian": {flag: 'hu',cs:'4221',as:'4729'},
"Italian": {flag: 'it',cs:'4348',as:'4485'},
"Latvian": {flag: 'lv',cs:'4674',as:''},
"Lithuanian": {flag: 'lt',cs:'4673',as:''},
"Norwegian": {flag: 'no',cs:'4356',as:'4533'},
"Polish": {flag: 'pl',cs:'4351',as:'4512'},
"Spanish": {flag: 'es',cs:'4346',as:'4524'},
"Swedish": {flag: 'se',cs:'4349',as:'4554'},
"Portuguese (PT)": {flag: 'pt',cs:'4347',as:'4522'},
"Romanian": {flag: 'ro',cs:'4675',as:''},
"Turkish": {flag: 'tr',cs:'4355',as:'4530'},
},
"Cantonese": {flag: 'cn',cs:'',as:''},
"Chinese": {flag: 'cn',cs:'4353',as:'4752'},
"English": {flag: 'us',cs:'4164',as:'4482'},
"Estonian": {flag: 'ee',cs:'4669',as:''},
"Hebrew": {flag: 'il',cs:'4670',as:'4727'},
"Icelanding": {flag: 'is',cs:'4672',as:''},
"Indonesia": {flag: 'id',cs:'4368',as:'4713'},
"Japanese": {flag: 'jp',cs:'4265',as:'4717'},
"Korean": {flag: 'kp',cs:'4366',as:''},
"Malay": {flag: 'ms',cs:'4367',as:''},
"Portuguese (BR)": {flag: 'br',cs:'4552',as:'4781'},
"Russian": {flag: 'ru',cs:'4350',as:'4559'},
"Serbian": {flag: 'rs',cs:'4677',as:''},
"Slovenian": {flag: 'si',cs:'4676',as:''},
"Tagalog": {flag: '',cs:'4678',as:''},
"Thai": {flag: 'th',cs:'4354',as:'4555'},
"Vietnamese": {flag: 'vn',cs:'4679',as:''}
};
$.each( langs, function( key, l ) {
//console.log('length L:',Object.keys(l).length,l);
if(Object.keys(l).length > 3)
{
if(key == "Europe") flag = 'eu';
else flag = '';
html = '<li class="dropdown-submenu">';
html += ' <a href="#"><span class="flag flag-'+flag+'"></span> '+key+'</a>';
html += ' <ul class="dropdown-menu">';
$.each( l, function( key2, l2 ) {
console.log(key2,l2);
if(l2.cs.length < 8)
l2.cs = '7108' + l.cs;
if(l2.as.length < 8 && l2.as.length > 0)
l.as = '7108' + l.as;
if(l2.as !== '')
{
html += '<li class="dropdown-submenu">';
html += ' <a href="tel:'+l.cs+'"><span class="flag flag-'+l2.flag+'"></span> '+key2+'</a>';
html += ' <ul class="dropdown-menu">';
html += ' <li><a href="tel:'+l2.cs+'">CS: '+l2.cs.substring(4,8)+'</a></li>';
html += ' <li><a href="tel:'+l2.as+'">AS: '+l2.as.substring(4,8)+'</a></li>';
html += ' </ul>';
html += '</li>';
}
else
{
html += '<li data-lang="'+l2.flag+'">';
html += ' <a href="tel:'+l2.cs+'"><span class="flag flag-'+l2.flag+'"></span> '+key2+'</a>';
html += '</li>';
}
});
$('#btb_lang_list').append(html);
}
else
{
if(l.cs.length < 8)
l.cs = '7108' + l.cs;
if(l.as.length < 8 && l.as.length > 0)
l.as = '7108' + l.as;
if(l.as !== '')
{
html = '<li class="dropdown-submenu">';
html += ' <a href="tel:'+l.cs+'"><span class="flag flag-'+l.flag+'"></span> '+key+'</a>';
html += ' <ul class="dropdown-menu">';
html += ' <li><a href="tel:'+l.cs+'">CS: '+l.cs.substring(4,8)+'</a></li>';
html += ' <li><a href="tel:'+l.as+'">AS: '+l.as.substring(4,8)+'</a></li>';
html += ' </ul>';
html += '</li>';
}
else
{
html = '<li data-lang="'+l.flag+'">';
html += ' <a href="tel:'+l.cs+'"><span class="flag flag-'+l.flag+'"></span> '+key+'</a>';
html += '</li>';
}
$('#btb_lang_list').append(html);
}
});
答案 0 :(得分:0)
IEnumerable
using System.Linq;
ToArray
。实施IEnumerable
是微不足道的。只需从列表中的根节点yield return Node.data;
开始,然后移动到下一个节点。冲洗并重复,直到下一个节点为null
。
答案 1 :(得分:0)
为您的班级添加新方法。
public class MyLinkedList
{
... keep existing methods here ...
public int[] ToArray() {
var result = new int[count]();
var index = 0;
var node = headnode;
while (node != null) {
result[index] = node.data;
node = node.next;
}
return result;
}
答案 2 :(得分:0)
public class MyLinkedList {
...
public Node[] ToArray() {
// You´ve got pre-computed count - let´s use it
Node[] result = new Node[count];
Node node = headNode;
for (int i = 0; i < result.Length; ++i) {
result[i] = node;
node = node.next;
}
return result;
}
}
P.S。 公共字段(例如next
类中的Node
是不良做法。将它们转换为属性
答案 3 :(得分:0)
正如@Will建议的那样,我会在你的班级实施IEnumerable<int>
。这将带您进入LINQ的强大世界,在这里您可以将链接列表转换为数组,列表或仅过滤节点:
public class MyLinkedList : IEnumerable<int>
{
// your code here
public IEnumerator<int> GetEnumerator()
{
Node current = headNode;
while(current != null)
{
yield return current.data;
current = current.next;
}
}
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
现在您可以在课程中使用所有LINQ扩展名:
MyLinkedList myList = new MyLinkedList();
myList.Add(10);
myList.Add(15);
myList.Add(20);
int[] array = myList.ToArray();
List<int> list = myList.ToList();
// items above 13
var items = myList.Where(i => i > 13);