23.Merge-k-Sorted-Lists

23. Merge k Sorted Lists

题目地址

题目描述

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
Example:
Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6 Given a string s, cut s into some substrings such that every substring is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.

代码

Approach 1: Divide & Conquer

class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
if (lists.size() == 0) return null;
return mergeHelper(lists, 0, lists.size() - 1);
}
private ListNode mergeHelper(List<ListNode> lists, int start, int end) {
if (start == end) return lists.get(start);
int mid = start + (end - start) / 2;
ListNode left = mergeHelper(lists, start, mid);
ListNode right = mergeHelper(lists, mid + 1, end);
return mergeTwoLists(left, right);
}
private ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
tail.next = list1;
tail = list1;
list1 = list1.next;
} else {
tail.next = list2;
tail = list2;
list2 = list2.next;
}
}
if (list1 != null) {
tail.next = list1;
} else {
tail.next = list2;
}
return dummy.next;
}
}
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
return partition(lists, 0, lists.length - 1);
}
private ListNode partition(ListNode[] lists, int start, int end) {
if (start == end) return lists[start];
if (start < end) {
int mid = start + (end - start) / 2;
ListNode node1 = partition(lists, start, mid);
ListNode node2 = partition(lists, mid + 1, end);
return merge(node1, node2);
} else {
return null;
}
}
private ListNode merge(ListNode node1, ListNode node2) {
if (node1 == null) return node2;
if (node2 == null) return node1;
if (node1.val < node2.val) {
node1.next = merge(node1.next, node2);
return node1;
} else {
node2.next = merge(node1, node2.next);
return node2;
}
}
}

Approach 2: PriorityQueue

class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
if (lists == null || lists.size() == 0) return null;
Queue<ListNode> heap = new PriorityQueue<ListNode>((a, b) -> a.val - b.val);
for (int i = 0; i < lists.size(); i++) {
if (lists.get(i) != null) {
heap.add(lists.get(i));
}
}
ListNode dummy = new ListNode(0);
ListNode head = dummy;
while (!heap.isEmpty()) {
ListNode node = heap.poll();
head.next = node;
head = node;
if (node != null) {
heap.add(node.next);
}
}
return dummy.next;
}
}

Approach 3: Merge two by two

class Soluution {
public ListNode mergeKLists(List<ListNode> Lists) {
if (lists == null || lists.size() == 0) return null;
while (lsits.size() > 1) {
List<ListNode> new_lists = new ArrayList<ListNode>();
for (int i = 0; i + 1 < lists.size(); i += 2) {
ListNode merged_list = merge(lists.get(i), lists.get(i + 1));
new_lists.add(merged_list);
}
if (lists.size() % 2 == 1) {
new_lists.add(lists.get(lists.size() - 1));
}
lists = new_lists;
}
return lists.get(0);
}
private ListNode merge(ListNode a, ListNode b) {
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while (a != null && b != null) {
if (a.val < b.val) {
tail.next = a;
a = a.next;
} else {
tail.next = b;
b = b.next;
}
tail = tail.next;
}
if (a != null) {
tail.next = a;
} else {
tail.next = b;
}
return dummy.next;
}
}

Approach 5: Brute Force

Intuition
  • Traverse all the linked lists and collect the values of the nodes into an array.
  • Sort and iterate over this array to get the proper value of nodes.
  • Create a new sorted linked list and extend it with the new nodes.
  • Time complexity : O_(_N_log_N) where N is the total number of nodes.
    • Collecting all the values costs O_(_N) time.
    • A stable sorting algorithm costs O_(_N_log_N) time.
    • Iterating for creating the linked list costs O_(_N) time.
  • Space complexity : O(N).
    • Sorting cost O(N) space (depends on the algorithm you choose).
    • Creating a new linked list costs O(N) space.
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) return null;
List<Integer> list = new ArrayList<Integer>();
for(int i = 0; i < lists.length; i++) {
ListNode node = lists[i];
while (node != null) {
list.add(node.val);
node = node.next;
}
}
ListNode dummy = new ListNode(0);
/// sort list and append to the dummy
return dummy;
}
}