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/*
* 23. Merge k Sorted Lists
* Target: Given n pairs of parentheses, generate all combinations of well-formed parentheses.
* Difficulty：Medium
* Classification：String, Backtracking
*/
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
/*
* Solution 1
* 2019-06-15 Runtime: 7 ms
* Algorithm: => Brute Force.
* Time complexity: O(n*logn). Space complexity: O(n).
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
List<Integer> nodes = new ArrayList<>();
ListNode head = new ListNode(0);
ListNode point = head;
for (ListNode l : lists){
while (l != null){
nodes.add(l.val);
l = l.next;
}
}
Collections.sort(nodes);
for (int k : nodes) {
point.next = new ListNode(k);
point = point.next;
}
return head.next;
}
}
/*
* Solution 2
* 2019-06-15 Runtime: 322 ms
* Algorithm: => Compare one by one.
* Time complexity: O(k*n). Space complexity: O(1).
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
int minIndex = 0;
ListNode head = new ListNode(0);
ListNode cur = head;
while(true){
boolean isBreak = true;
int minVal = Integer.MAX_VALUE;
for (int i = 0; i < lists.length; i++){
if (lists[i] != null) {
if (lists[i].val < minVal){
minIndex = i;
minVal = lists[i].val;
}
isBreak = false;
}
}
if (isBreak){
break;
}
cur.next = lists[minIndex];
cur = cur.next;
lists[minIndex] = lists[minIndex].next;
}
cur.next = null;
return head.next;
}
}
/*
* Solution 3
* 2019-06-15 Runtime: 5 ms
* Algorithm: => Optimize solution 2 by Priority Queue.
* Time complexity: O(nlog(k)). Space complexity: O(k).
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
Comparator<ListNode> cmp = new Comparator<ListNode>() {
@Override
public int compare(ListNode l1, ListNode l2){
return (l1.val - l2.val);
}
};
Queue<ListNode> q = new PriorityQueue<ListNode>(cmp);
for (ListNode l : lists) {
if (l != null) {
q.add(l);
}
}
ListNode head = new ListNode(0);
ListNode cur = head;
while (!q.isEmpty()){
cur.next = q.poll();
cur = cur.next;
ListNode next = cur.next;
if (next != null){
q.add(next);
}
}
return head.next;
}
}
/*
* Solution 4
* 2019-06-15 Runtime: 302 ms
* Algorithm: => Merge lists one by one.
* Time complexity: O(k*n). Space complexity: O(1).
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 != null && l2 != null){
if (l1.val > l2.val){
ListNode temp = l1;
l1 = l2;
l2 = temp;
}
l1.next = mergeTwoLists(l1.next, l2);
}
return (l1 != null) ? l1 : l2;
}
public ListNode mergeKLists(ListNode[] lists) {
if (lists.length == 0){
return null;
}
if(lists.length == 1){
return lists[0];
}
ListNode head = mergeTwoLists(lists[0], lists[1]);
for (int i = 2; i < lists.length; i++){
head = mergeTwoLists(head, lists[i]);
}
return head;
}
}
/*
* Solution 5
* 2019-06-16 Runtime: 2 ms
* Algorithm: => Merge with Divide And Conquer.
* Time complexity: O(nlog(k)). Space complexity: O(1).
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 != null && l2 != null){
if (l1.val > l2.val){
ListNode temp = l1;
l1 = l2;
l2 = temp;
}
l1.next = mergeTwoLists(l1.next, l2);
}
return (l1 != null) ? l1 : l2;
}
public ListNode mergeKLists(ListNode[] lists) {
if (lists.length == 0){
return null;
}
int point = 1;
while (point < lists.length){
for (int i = 0; i + point < lists.length; i = i + point * 2){
lists[i] = mergeTwoLists(lists[i], lists[i + point]);
}
point *= 2;
}
return lists[0];
}
}