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import java.util.Comparator;

import java.util.PriorityQueue;

/**

* https://leetcode.com/articles/merge-k-sorted-list/

*/

public class MergeKSortedList {

// 耗时19ms

// 时间复杂度为O(knlgn)，空间复杂度O(k)

/**

* 这里要注意lists中可能有node为null

*/

public ListNode mergeKLists(ListNode[] lists) {

ListNode dummy = new ListNode(0), cur = dummy;

PriorityQueue<ListNode> queue = new PriorityQueue<>(new Comparator<ListNode>() {

@Override

public int compare(ListNode node1, ListNode node2) {

return node1.val - node2.val;

}

});

for (ListNode node : lists) {

if (node != null) {

queue.offer(node);

}

}

while (!queue.isEmpty()) {

ListNode node = queue.poll();

cur.next = node;

cur = cur.next;

if (node.next != null) {

queue.offer(node.next);

}

}

return dummy.next;

}

/**

* 第二种方法

* 时间复杂度O(Nlgk)，空间O(l)

* 耗时10ms

*/

public ListNode mergeKLists2(ListNode[] lists) {

return helper(lists, 0, lists.length - 1);

}

private ListNode helper(ListNode[] lists, int start, int end) {

if (start > end) {

return null;

}

if (start == end) {

return lists[start];

}

int mid = start + (end - start) / 2;

ListNode l1 = helper(lists, start, mid);

ListNode l2 = helper(lists, mid + 1, end);

return mergeTwoLists(l1, l2);

}

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

ListNode dummy = new ListNode(0);

ListNode p = l1, q = l2, cur = dummy;

for ( ; p != null && q != null; ) {

if (p.val < q.val) {

cur.next = p;

p = p.next;

} else {

cur.next = q;

q = q.next;

}

cur = cur.next;

}

cur.next = p != null ? p : q;

return dummy.next;

}

}