Divide-and-conquer algo, merge list in pairs, then again, until only one list is left.
Created
April 9, 2023 17:27
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Given an array of k sorted linked lists, your task is to merge them into a single sorted list.
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function mergeTwoLists(list1, list2) { | |
let a = list1.head; | |
let b = list2.head; | |
let ll = null; | |
let tail = null; | |
while (a !== null || b !== null) { | |
let nextNode = null; | |
if (a !== null && b !== null) { | |
if (a.data <= b.data) { | |
nextNode = a; | |
a = a.next; | |
} else { | |
nextNode = b; | |
b = b.next; | |
} | |
} else { | |
if (b === null && a !== null) { | |
nextNode = a; | |
a = a.next; | |
} else if (a === null && b !== null) { | |
nextNode = b; | |
b = b.next; | |
} | |
} | |
if (tail === null) { | |
ll = new LinkedList(); | |
ll.createLinkedList([nextNode.data]); | |
tail = ll.head; | |
} else { | |
tail.next = new LinkedListNode(nextNode.data); | |
tail = tail.next; | |
} | |
} | |
return ll; | |
} | |
export function mergeKLists(lists) { | |
if (lists.length === 0) { | |
return; | |
} | |
let step = 1; | |
while (step < lists.length) { | |
for (let i = 0; i < lists.length - step; i = i + step * 2) { | |
lists[i] = mergeTwoLists(lists[i], lists[i + step]); | |
} | |
step *= 2; | |
} | |
return lists[0].head; | |
} |
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