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题目

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

方案：

/**
 * 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];
            ListNode[] l2 = new ListNode[lists.length- mid];
            for (int i = 0 ;i<mid;i++){
                l1[i] = lists[i];
            }
            for(int i = mid;i<lists.length;i++){
                l2[i-mid] = lists[i];
            }
            return mergeTwoLists(mergeKLists(l1),mergeKLists(l2));
        }

        public ListNode mergeTwoLists(ListNode l1,ListNode l2) {
            if (l1==null){
                return l2;
            }else if (l2==null){
                return l1;
            }
            ListNode head, tail;
            //第一个节点
            if (l1.val > l2.val) {
                head = l2;
                l2=l2.next;
            } else {
                head = l1;
                l1=l1.next;
            }
            tail = head;
            while (l1 != null || l2 != null) {
                if (l1 == null && l2 != null) {
                    tail.next = l2;
                    l2 = l2.next;
                } else if (l1 != null && l2 == null) {
                    tail.next = l1;
                    l1 = l1.next;
                }else if (l1.val > l2.val) {
                    tail.next = l2;
                    l2 = l2.next;
                }else if (l1.val <= l2.val) {
                    tail.next = l1;
                    l1 = l1.next;
                }
                tail = tail.next;
            }
            return head;
        }

    }

复杂度分析

• 时间复杂度：O(n²)
• 空间复杂度：O(1)