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;
}
}