Please try to complete the following cloze:
/* Create a Queue, head is head0, tail is tail0 */
function IntList(head0, tail0){
this.head = head0 || 0;
this.tail = tail0 || null;
}
/* Return an IntList containing all the numbers in the array*/
IntList.list = function(__args){
var sentinel = new IntList(),
len = __args.length,
p = sentinel;
for(var i = 0; i < len; i ){
p.tail = new IntList(__args[i]);
p = p.tail ;
}
return sentinel.tail;
};
/* Return the string representation of the object*/
IntList.prototype.toString = function(){
var temp = "";
temp = "[";
for(var L = this; L !== null; L = L.tail){
temp = temp " " L.head;
}
temp = " ]";
return temp;
};
/**Returns an IntList containing IntList A and IntList B,
* where the elements of B are after A. The new keyword cannot be used.
*/
function dcatenate(A, B){
/* Complete function*/
}
/**Returns a new IntList with length len,
* starting with the #start element (where #0 is the first element),
* cannot change L.
*/
function sublist(L, start, len){
/* Complete function*/
}
This is a linked list question written in Javascript. Since linked lists have relatively complex reference operations, they can be used to examine the understanding of Javascript references. Comes with a simple test case:
/* Test dcatenate and sublist Is the function correct*/
function test(){
var A = IntList.list([4,6,7,3,8]),
B = IntList.list([3,2, 5,9]);
dcatenate(A, B);
if(A.toString() === "[ 4 6 7 3 8 3 2 5 9 ]"){
alert(" The dcatenate function is correct. ");
}else{
alert("The dcatenate function is wrong.");
}
var L = IntList.list([3,4,5,2,6 ,8,1,9]),
result = sublist(L, 3, 3);
if(result.toString() === "[ 2 6 8 ]"){
alert( "The sublist function is correct.");
}else{
alert("The sublist function is correct.");
}
}
Javascript reference?
In fact, when assigning a reference type instance to a variable, the variable holds a reference to the instance:
var temp = new Object();
This behavior is very consistent with its name, reference type, and its instances are of course referenced.
And when the variable is assigned to another variable, it is actually just a copy of the reference:
var temp2 = temp;
So although from the definition: temp2 = temp, they are not directly related. For example, the reference of temp is modified:
var temp = {
name: "temp"
};
var temp2 = temp;
temp = {
name: "not temp2"
};
temp === temp2; //false
Of course, if we modify only the instance pointed to by the pointer, then temp2 is still equal to temp:
var temp = {
name: " temp"
};
var temp2 = temp;
temp.name = "also temp2";
temp === temp2; //true
What is IntList? Let’s analyze the picture below:
Create two empty variables, so L and Q in the picture on the right are empty. Create a new IntList with a head of 3 and an empty tail, and assign the value referenced by L to Q, so both L and Q point to this new IntList. Q points to a newly created IntList whose head is 42 and whose tail is empty. The pointer of Q is assigned to L.tail, so that the two IntLists are nested.
It can be seen that IntList is a data structure that achieves multiple nesting through pointers, which is called a Linked List.
1. Create two empty variables, so L and Q in the picture on the right are empty.
2. Create a new IntList with a head of 3 and an empty tail. Assign the value referenced by L to Q, so both L and Q point to this new IntList.
3.Q points to a newly created IntList whose head is 42 and whose tail is empty. Assign the pointer of Q to L.tail, so that the two IntLists are nested.
It can be seen that IntList is a data structure that achieves multiple nesting through pointers, which is called a Linked List.
IntList merge
We just need to point the tail of one to the other. In this way, the two IntLists are connected:
/**Returns an IntList containing IntList A and IntList B,
* where the elements of B are after A. The new keyword cannot be used.
*/
function dcatenate(A, B){
var p;
for(p = A; p != null; p = p.tail){
if(p.tail === null){
p.tail = B;
break;
}
}
return A
}
IntList interception
Since the question requires that the original IntList cannot be changed, we can only take out the data from the original IntList and reconstruct new data.
/**Returns a new IntList with length len,
* starting with the #start element (where #0 is the first element),
* cannot change L.
*/
function sublist(L, start, len){
var K,
P,
J;
var i = 0,
end = start len;
for(P = L; i <= end; P = P.tail, i ){
if(i < start){
continue;
}else if(i === start){
K = new IntList(P.head);
J = K;
}else if(i > start && i < end){
J.tail = new IntList(P.head);
J = J.tail;
}else if(i >= end){
break;
}
}
return K;
}
Thinking Questions
1. How do functions pass parameters when they are passed? For example, what is the citation process for the following code?
var obj = {
name: "anything"
};
function getName(__obj){
return __obj.name;
}
var name = getName(obj);