Scal dwie posortowane listy połączone
/*
This implementation demonstrates how to
merge two sorted lists into an aggregate
sorted one.
Let n be the number of elements of first
singly linked list and m the number of
elements of the second one.
Time complexity: O(n + m)
Space complexity: O(1)
*/
function mergeLinkedLists(headOne, headTwo) {
let ptr1 = headOne;
let ptr1Prev = null;
let ptr2 = headTwo;
while (ptr1 !== null && ptr2 !== null) {
// If smaller value belongs to first list
// advance to next node in first list
if (ptr1.value < ptr2.value) {
ptr1Prev = ptr1;
ptr1 = ptr1.next;
} else {
// Otherwise, advance to next node in second
if (ptr1Prev !== null) ptr1Prev.next = ptr2;
ptr1Prev = ptr2;
ptr2 = ptr2.next;
ptr1Prev.next = ptr1;
}
}
if (ptr1 === null) ptr1Prev.next = ptr2;
return headOne.value < headTwo.value ? headOne : headTwo;
}
class Node {
constructor(value) {
this.value = value;
this.next = null;
}
}
class LinkedList {
constructor() {
this.head = null;
}
// Add a value at end of list
addEnd(value) {
const node = new Node(value);
let curr = this.head;
if (curr == null) {
this.head = node;
return;
}
while (curr !== null && curr.next !== null) {
curr = curr.next;
}
curr.next = node;
}
}
const list1 = new LinkedList();
const list2 = new LinkedList();
const aggregateList = new LinkedList();
list1.addEnd(1);
list1.addEnd(4);
list1.addEnd(6);
list2.addEnd(2);
list2.addEnd(5);
aggregateList.head = mergeLinkedLists(list1.head, list2.head);
let temp = aggregateList.head;
const aggregateListContent = [];
while (temp !== null) {
aggregateListContent.push(temp.value + " ");
temp = temp.next;
}
// Below prints: 1 2 4 5 6
console.log("Aggregate list:", aggregateListContent.join(""));
Wissam