合并 k 个排序链表,返回合并后的排序链表。请分析和描述算法的复杂度。
示例:
输入:
[
1->4->5,
1->3->4,
2->6
]
输出: 1->1->2->3->4->4->5->6
PS:直接用PriorityQueue自动排序,改写一下compare方法。
/**
- 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.length == 0) { return null; } ListNode dummyHead = new ListNode(0); ListNode curr = dummyHead; PriorityQueue<ListNode> pq = new PriorityQueue<>(new Comparator<ListNode>() { @Override public int compare(ListNode o1, ListNode o2) { return o1.val - o2.val; } }); for (ListNode list : lists) { if (list == null) { continue; } pq.add(list); } while (!pq.isEmpty()) { ListNode nextNode = pq.poll(); curr.next = nextNode; curr = curr.next; if (nextNode.next != null) { pq.add(nextNode.next); } } return dummyHead.next;
}
}
PS:分治
/**
- 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.length == 0)
return null;
if(lists.length == 1)
return lists[0];
if(lists.length == 2){
return mergeTwoLists(lists[0],lists[1]);
}int mid = lists.length/2; ListNode[] l1 = new ListNode[mid]; for(int i = 0; i < mid; i++){ l1[i] = lists[i]; } ListNode[] l2 = new ListNode[lists.length-mid]; for(int i = mid,j=0; i < lists.length; i++,j++){ l2[j] = lists[i]; } return mergeTwoLists(mergeKLists(l1),mergeKLists(l2));
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;ListNode head = null; if (l1.val <= l2.val){ head = l1; head.next = mergeTwoLists(l1.next, l2); } else { head = l2; head.next = mergeTwoLists(l1, l2.next); } return head;
}
}
4
原文地址:https://blog.51cto.com/14073038/2471173
时间: 2024-11-07 14:28:22