反转链接列表..我的实现有什么问题

Question

基于下面的实现，我在反转链表时遇到了问题。我在这里是不是做错了什么？

class Node {
constructor(val) {
this.val = val;
this.next = null;
}
}
class SinglyLinkedList {
constructor() {
this.head = null;
this.length = 0;
}
push(val) {
var newNode = new Node(val);
var current = this.head; 
if (!this.head) 
this.head = newNode; 
else {
// iterate to the end of the 
// list 
while (current.next) { 
current = current.next; 
} 
// add node 
current.next = newNode; 
} 
this.length++;
return this; 
}
// reverse the list
reverse() {
var prev = null;
var curr = this.head;
while (curr !== null) {
var temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;
}
return this;
}
print() {
var arr = []
var current = this.head;
while(current) {
arr.push(current.val);
current = current.next; 
}
console.log(arr);
}
}

以下是我创建对象并推送一些节点时的实现

var list = new SinglyLinkedList();
list.push(1);
list.push(2);
list.push(3);
list.push(4);

每次我运行list.reverse()然后运行list.print()时，它只打印[1]而不打印[4,3,2,1]。

Answer 1

您正确地反转了节点之间的链接，但从未更改this.head指向的内容，因此它现在指向列表的末尾，而不是列表的前面。因此，当您调用print时，print将从最后一个节点开始，然后无处可去。

reverse() {
var prev = null;
var curr = this.head;
while (curr !== null) {
var temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;
}
this.head = prev; // <--- added
return this;
}

Answer 2

您尚未更新reverse方法中的head属性。只需在while循环之后添加this.head = prev;，我相信它应该可以工作。