Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
There are two ways, first use ArrayList to rewrite the linkedlist array, then sort it and use a ListNode to finalize it.
public ListNode mergeKLists(ListNode[] lists) {
ArrayList<Integer> list = new ArrayList<>();
for (ListNode l : lists) {
while (l != null) {
list.add(l.val);
l = l.next;
}
}
Collections.sort(list);
ListNode ans = new ListNode(0);
ListNode a = ans;
for (int i = 0; i < list.size(); i++) {
ListNode tmp = new ListNode(list.get(i));
a.next = tmp;
a = a.next;
}
return ans.next;
}
Second is to use a priorityQueue and change its comparator function.
知识点：Priority Queue (Use tree)
Two methods to add nodes, add() & offer(). Difference : add() throws an exception, offer() return false;
Two methods to get the first node, element() & peek(). Difference : element() throws an exception, peek() return null;
Two methods to get and remove the first node, remove() & poll(). Difference : remove() throws an exception, poll() return null;
public ListNode mergeKLists(ListNode[] lists) {
if (lists.length == 0 || lists == null) return null;
PriorityQueue<ListNode> queue = new PriorityQueue<ListNode>(lists.length, new Comparator<ListNode>() {
@Override
public int Comparator(ListNode n1, ListNode n2) {
if (n1.val < n2.val) {
return -1;
} else if (n1.val == n2.val) {
return 0;
} else {
return 1;
}
}
});
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
for (ListNode l : lists) {
if (l != null) {
queue.offer(l);
}
}
while (!queue.isEmpty()) {
tail.next = queu.poll();
tail = tail.next;
if (tail.next != null) {
queue.offer(tail.next);
}
}
return dummy.next;
}
原文地址：https://www.cnblogs.com/jjjiajia/p/8452456.html