给你一个链表数组,每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中,返回合并后的链表。
https://leetcode-cn.com/problems/merge-k-sorted-lists/
示例1:
示例2:
示例3:
提示:
Java解法
package sj.shimmer.algorithm.ten_2;
import sj.shimmer.algorithm.ListNode;
/**
* Created by SJ on 2021/2/9.
*/
class D16 {
public static void main(String[] args) {
ListNode n1 = ListNode.createNode(new int[]{1,4,5});
ListNode n2 = ListNode.createNode(new int[]{1,3,4});
ListNode n3 = ListNode.createNode(new int[]{2,6});
ListNode n4 = ListNode.createNode(new int[]{});
// System.out.println(mergeKLists(new ListNode[]{n1,n2,n3}));
// System.out.println(mergeKLists(new ListNode[]{}));
// System.out.println(mergeKLists(new ListNode[]{n4}));
System.out.println(mergeKLists2(new ListNode[]{n1,n2,n3}));
System.out.println(mergeKLists2(new ListNode[]{}));
System.out.println(mergeKLists2(new ListNode[]{n4}));
}
public static ListNode mergeKLists2(ListNode[] lists) {
ListNode preHead = new ListNode();
ListNode pre = preHead;
int minIndex = getMin(lists);
while (minIndex!=-1) {
ListNode node = lists[minIndex];
pre.next = node;
pre = pre.next;
lists[minIndex] = node.next;
minIndex = getMin(lists);
}
return preHead.next;
}
public static ListNode mergeKLists(ListNode[] lists) {
ListNode head = null;
int minIndex = getMin(lists);
if (minIndex==-1) {
return null;
}
head = lists[minIndex];
lists[minIndex] = head.next;
head.next = mergeKLists(lists);
return head;
}
public static int getMin(ListNode[] lists) {
ListNode min = null;
int minIndex = -1;
for (int i = 0; i < lists.length; i++) {
if (lists[i] == null) {
continue;
}
if (min == null||min.val>lists[i].val) {
min = lists[i];
minIndex = i;
}
}
return minIndex;
}
}
官方解
-
顺序合并
- 时间复杂度: O(k^2 *n)
- 空间复杂度:O(1)
-
分治合并
- 时间复杂度为O(kn×logk)。
- 空间复杂度:递归会使用到 O(logk)