双链表为什么不能删除第一个元素

Question

我在编程时非常“绿色”，我明天要带任务。它几乎完成，但有一个小问题。我不能删除第一个元素，并且如果在尝试删除第一个元素之后，我输入 一个新的元素在它的位置mmm ....让我们只是说我输入了无数其中的许多元素。我似乎无法找到问题所在。在此先感谢

program dvipsar; 

type duomenys = integer; 
    sarasas = ^elementas; 
    elementas = record 
     info: duomenys; 
     anks: sarasas; 
     tolsn: sarasas 
    end; 


procedure sukurtiTuscia(var s: sarasas); {creates empty list} 
begin 
    s := nil 
end; 

function tuscias(s: sarasas): boolean; {checks if list is empty} 
begin 
    tuscias := s = nil 
end; 

function elmSk(s: sarasas): integer;  {counts elements} 
    var kiek: integer; 
begin 
    kiek := 0; 
    while s <> nil do 
    begin 
    kiek := kiek + 1; 
    s := s^.tolsn 
    end; 
    elmSk := kiek 
end; 

function gautiRodN(s: sarasas; n:integer): sarasas;  {Arrow to n-th element} 
    var i: integer; 
begin 
    i := 1; 
    while (s <> nil) and (i<n) do 
    begin 
    i := i + 1; 
    s := s^.tolsn 
    end; 
    if i = n then gautiRodN := s 
    else gautiRodN := nil 
end; 

function gautiN(s: sarasas; n:integer): duomenys;   {gets n-th element data} 
    var elem: sarasas; 
begin 
    elem := gautiRodN(s,n); 
    if elem <> nil then gautiN := elem^.info 
end; 

procedure iterptiPries(s:sarasas; n: integer; duom: duomenys);  {adds new element before n-th element} 
    var nElem: sarasas; 
     naujas: sarasas; 
begin 
    nElem := gautiRodN(s,n); 
    if nElem <> nil then begin 
    new (naujas); 
    naujas^.info := duom; 
    naujas^.tolsn := nElem; 
    naujas^.anks := nElem^.anks; 
    if nElem^.anks <> nil then nElem^.anks^.tolsn := naujas; 
    nElem^.anks := naujas; 
end 
end; 

procedure panaikintiN(s: sarasas; n: integer);   {removes element from n-th place} 
    var nElem: sarasas; 
begin 
    nElem := gautiRodN(s,n); 
    if nElem <> nil then begin 
    if nElem^.anks <> nil then nElem^.anks^.tolsn := nElem^.tolsn; 
    if nElem^.tolsn <> nil then nElem^.tolsn^.anks := nElem^.anks; 
    dispose(nElem); 
    end; 
end; 

function rasti(s: sarasas; duom: duomenys): sarasas;  {finds element} 
begin 
    while (s <> nil) and (s^.info <> duom) do s := s^.tolsn; 
    rasti := s 
end; 

procedure spausdinti(s: sarasas);    {prints list} 
begin 
    while (s <> nil) do begin 
    write(s^.info,' '); 
    s := s^.tolsn 
    end; 
    writeln 
end; 

procedure panaikintiP(var s: sarasas); {removes first element} 
    var pirmas: sarasas; 
begin 
    pirmas := s; 
    s := s^.tolsn; 
    dispose (pirmas) 
end; 

procedure panaikinti(var s: sarasas);      {deletes list} 
begin 
    while s <> nil do panaikintiP(s) 
end; 


procedure prideti(var s: sarasas; duom: duomenys);  {add element at the end of the list} 
    var kiek: integer; 
     paskutinis,naujas: sarasas; 
begin 
    kiek := elmSk(s); 
    paskutinis := gautiRodN(s,kiek); 
    new(naujas); 
    naujas^.info := duom; 
    naujas^.tolsn := nil; 
    naujas^.anks := paskutinis; 
    if paskutinis <> nil then paskutinis^.tolsn := naujas 
    else s := naujas 
end; 

procedure menu; 
begin 
    writeln; 
    writeln; 
    writeln ('1 Creat a list'); 
    writeln ('2 Count the elements'); 
    writeln ('3 Check if list is empty'); 
    writeln ('4 Print an element'); 
    writeln ('5 Print the list'); 
    writeln ('6 Remove an element') 
    writeln ('7 Add an element'); 
    writeln ('8 Search in the list'); 
    writeln; 
    writeln ('0 End'); 
    writeln; 
    writeln; 

end; 
var s: sarasas; 
    i,j: integer; 
    t: sarasas; 
    c: char; 
    veiksmas: integer; 
begin 
    sukurtiTuscia(s); 
    repeat 
    menu; 
    write('Input action number : '); 
    readln(veiksmas); 
    case veiksmas of 
     1: 
     repeat 
      write('input a number which you want to add to the list: '); 
      readln(i); 
      prideti(s,i); 
      write('Add new number? (t/n)? '); 
      read(c); 
     until (c='N') or (c='n'); 
     2: writeln ('List is not empty: ',elmSk(s)); 
     3: if tuscias(s) then writeln ('List is empty') 
         else writeln ('List is not empty'); 
     4: begin 
     write ('Which element to print?: '); 
     readln(i); 
     writeln(i,'-th list element?: ',gautiN(s,i)); 
     end; 
     5: spausdinti(s); 
     6: begin 
     write ('which element to remove?: '); 
     readln(i); 
     panaikintiN(s,i); 
     end; 
     7:begin 
     write ('What to add to the list?: '); 
     readln(i); 
     write ('Before which element?: '); 
     readln(j); 
     iterptiPries(s,j,i); 
     end; 
     8: begin 
     write ('What element to look for?: '); 
     readln(i); 
     t := rasti(s,i); 
     if t <> nil then writeln (i, ' exists in the list') 
        else writeln (i, ' does not exists in the list'); 
     end; 
     0: writeln ('Ending'); 
    else writeln('Incorrect action'); 
    end; 
    until veiksmas = 0; 
    panaikinti(s);     {deletes list} 
end. 

Answer 1

http://en.wikipedia.org/wiki/Doubly_linked_list#Removing_a_node

节点的去除比插入更容易，但需要特殊处理，如果要删除的节点是firstNode或lastNode：

function remove(List list, Node node) 
    if node.prev == null 
     list.firstNode := node.next 
    else 
     node.prev.next := node.next 
    if node.next == null 
     list.lastNode := node.prev 
    else 
     node.next.prev := node.prev 
    destroy node 

一个微妙的后果上面的过程是删除列表的最后一个节点将firstNode和lastNode都设置为null，因此它会正确处理从一个元素列表中删除最后一个节点。请注意，我们也不需要单独的“removeBefore”或“removeAfter”方法，因为在双向链表中，我们可以只使用“remove（node.prev）”或“remove（node.next）”，其中这些都是有效的。

这也假设被删除的节点保证存在。

如果节点不存在于此列表中，那么将需要一些错误处理。