输入:5-> 9-> 8-> 3-> 1-> 7
预期输出:7-> 1-> 3-> 8-> 9-> 5
问题:
当我显示反向链接列表时,结果为5。这是一个问题,因为它应该是尾巴而不是头部。其余节点在显示孔中都缺失。
问题:
代码库是否存在阻止从头到尾遍历并更改指针以反转列表的问题?
代码:
LinkedList:
class LinkedList {
constructor() {
this.head = null;
this.size = 0;
}
insertFirst(item) {
if (this.head !== null) {
const newHead = new _Node(item);
let oldHead = this.head;
oldHead.prev = newHead;
newHead.next = oldHead;
this.head = newHead;
} else {
this.head = new _Node(item, this.head);
}
this.size++;
}
insertLast(item) {
if (!this.head) {
this.insertFirst(item);
} else {
let tempNode = this.head;
while (tempNode.next !== null) {
tempNode = tempNode.next;
}
tempNode.next = new _Node(item, null, tempNode);
}
this.size++
}
}
module.exports = LinkedList;
主要:
const LinkedList = require("./LinkedLists");
const { reverse } = require("./Reverse");
const { display } = require("./Supplemental");
function main() {
let SLL = new LinkedList();
SLL.insertFirst(5);
SLL.insertLast(9);
SLL.insertLast(8);
SLL.insertLast(3);
SLL.insertLast(1);
SLL.insertLast(7);
reverse(SLL);
display(SLL);
return SLL;
}
console.log(main());
反向:
reverse = (SLL) => {
let curr = SLL.head
if (!curr) {
return;
}
if (!curr.next) {
SLL.head = curr;
return;
}
let tmp = reverse(curr.next);
curr.next.next = curr;
curr.next = null;
return tmp;
}
module.exports = { reverse };
显示:
display = (SLL) => {
let currentNode = SLL.head;
if (!SLL.head) {
return null;
}
while (currentNode !== null) {
console.log(currentNode.value);
currentNode = currentNode.next;
}
return;
};
答案 0 :(得分:0)
有人可以告诉我return tmp
在Reverse.js文件中做什么吗?
(1)从Main.js中删除了显示
(2)编辑的Main.js:
const LinkedList = require("./LinkedLists");
const { reverse } = require("./Reverse");
const { display } = require("./Supplemental");
function main() {
let SLL = new LinkedList();
SLL.insertFirst(5);
SLL.insertLast(9);
SLL.insertLast(8);
SLL.insertLast(3);
SLL.insertLast(1);
SLL.insertLast(7);
const result = reverse(SLL.head);
console.log(result);
return SLL;
}
return main();
(3)编辑的Reverse.js:
reverse = (curr, prev = null) => {
if (!curr) {
return prev;
}
let tmp = reverse(curr.next, curr);
const temp = curr.next;
curr.next = prev;
curr.prev = temp;
return tmp;
}
module.exports = { reverse };