题目描述
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
解题思路
使用Priority Queue。这样保持每次取出的节点中是当前最小的，依次加入新的链表，从而得到合并的结果。
这个算法每个元素要读取一次，即是k*n次，然后每次读取元素要把新元素插入堆中要logk的复杂度，所以总时间复杂度是O(nklogk)。空间复杂度是堆的大小，即为O(k)。
Java代码实现
/**
* Definition for singly-linked list.
* 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;
}
// new PriorityQueue<>(lists.length, Comparator.comparingInt(o -> o.val));
PriorityQueue<ListNode> pq = new PriorityQueue<>(lists.length, (o1, o2) -> {
return Integer.compare(o1.val, o2.val);
});
ListNode dummy = new ListNode(0);
ListNode walk = dummy;
for (ListNode node : lists) {
if (node != null) {
pq.offer(node);
}
}
while (!pq.isEmpty()) {
walk.next = pq.poll();
walk = walk.next;
if (walk.next != null) {
pq.add(walk.next);
}
}
return dummy.next;
}
}
