[Algorithm] Reverse a linked list
It helps to understands how recursive calls works.
function Node(val) { return { val, next: null }; } function LinkedList() { return { head: null, tail: null, add(val) { const node = new Node(val); if (!this.head) { this.head = node; this.tail = node; return node; } this.tail.next = node; this.tail = node; return node; }, // 1 - -2 -- x-- x reverse() { const helper = node => { if (!node.next) { this.head = node; return; } helper(node.next); // after helper call ends // node is three // node.next is four // swap thre and four and point three next to null let temp = node.next; temp.next = node; node.next = null; }; return helper(this.head); } }; } const l = new LinkedList(); l.add("one"); l.add("two"); l.add("three"); l.add("four"); l.reverse(); console.log(l.head) // {"val":"four","next":{"val":"three","next":{"val":"two","next":{"val":"one","next":null}}}}
So for our 'helper' function, when calling it, it stop there until when reach the end.
one |
two |
three |
four |
v
helper()
four |
three |
tow |
one v
To reverse the linked list, everytime we just swap last two node, then set node.next = null.
Here we also should the apporach to using iteration:
function Node(val) { return { val, next: null }; } function LinkedList() { return { head: null, tail: null, add(val) { const node = new Node(val); if (!this.head) { this.head = node; this.tail = node; return node; } this.tail.next = node; this.tail = node; return node; }, reverse() { let current = this.head; let prev = null; while(current) { let next = current.next; current.next = prev; prev = current; current = next; } this.head = prev; } }; } const l = new LinkedList(); l.add("one"); l.add("two"); l.add("three"); l.add("four"); l.reverse(); console.log(l.head) // {"val":"four","next":{"val":"three","next":{"val":"two","next":{"val":"one","next":null}}}}