* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
4.7 KiB
4.7 KiB
id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
587d8251367417b2b2512c63 | 从链接列表中删除元素 | 1 | remove-elements-from-a-linked-list |
--description--
链接列表的任何实现所需的下一个重要方法是remove
方法。此方法应将要删除的元素作为参数,然后搜索列表以查找并删除包含该元素的节点。每当我们从链表中删除一个节点时,重要的是我们不要意外地孤立列表的其余部分。回想一下,每个节点的next
属性都指向列表中跟随它的节点。如果我们删除中间元素,比如说,我们要确保我们从该元素的前一个节点的next
属性到中间元素的next
属性(这是列表中的下一个节点)的连接!这可能听起来真的很混乱,所以让我们回到康加线的例子,这样我们就有了一个很好的概念模型。想象自己在康加舞线上,直接在你面前的人离开了这条线。刚离开生产线的人不再将手放在任何人身上 - 而且你不再把手放在离开的人身上。你向前走,把你的手放在你看到的下一个人身上。如果我们要删除的元素是head
元素,我们将head
重新分配给链表的第二个节点。
--instructions--
编写一个remove
方法,该方法接受一个元素并将其从链表中删除。注意每次从链接列表中删除元素时,列表的length
应减少一。
--hints--
您的LinkedList
类应该有一个remove
方法。
assert(
(function () {
var test = new LinkedList();
return typeof test.remove === 'function';
})(),
'Your <code>LinkedList</code> class should have a <code>remove</code> method.'
);
remove
第一个节点时, remove
方法应重新分配head
到第二个节点。
assert(
(function () {
var test = new LinkedList();
test.add('cat');
test.add('dog');
test.remove('cat');
return test.head().element === 'dog';
})(),
'Your <code>remove</code> method should reassign <code>head</code> to the second node when the first node is removed.'
);
对于每个删除的节点,您的remove
方法应该将链表的length
减少一个。
assert(
(function () {
var test = new LinkedList();
test.add('cat');
test.add('dog');
test.remove('cat');
return test.size() === 1;
})(),
'Your <code>remove</code> method should decrease the <code>length</code> of the linked list by one for every node removed.'
);
您的remove
方法应该将已删除节点的上next
节点的引用重新分配给已删除节点的next
引用。
assert(
(function () {
var test = new LinkedList();
test.add('cat');
test.add('dog');
test.add('kitten');
test.remove('dog');
return test.head().next.element === 'kitten';
})(),
'Your <code>remove</code> method should reassign the reference of the previous node of the removed node to the removed node's <code>next</code> reference.'
);
--seed--
--seed-contents--
function LinkedList() {
var length = 0;
var head = null;
var Node = function(element){
this.element = element;
this.next = null;
};
this.size = function(){
return length;
};
this.head = function(){
return head;
};
this.add = function(element){
var node = new Node(element);
if(head === null){
head = node;
} else {
var currentNode = head;
while(currentNode.next){
currentNode = currentNode.next;
}
currentNode.next = node;
}
length++;
};
this.remove = function(element){
// Only change code below this line
// Only change code above this line
};
}
--solutions--
function LinkedList() {
var length = 0;
var head = null;
var Node = function(element){
this.element = element;
this.next = null;
};
this.size = function(){
return length;
};
this.head = function(){
return head;
};
this.add = function(element){
var node = new Node(element);
if(head === null){
head = node;
} else {
var currentNode = head;
while(currentNode.next){
currentNode = currentNode.next;
}
currentNode.next = node;
}
length++;
};
this.remove = function(element){
if (head === null) {
return;
}
var previous;
var currentNode = head;
while (currentNode.next !== null && currentNode.element !== element) {
previous = currentNode;
currentNode = currentNode.next;
}
if (currentNode.next === null && currentNode.element !== element) {
return;
}
else if (previous) {
previous.next = currentNode.next;
} else {
head = currentNode.next;
}
length--;
};
}