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题目

输入k个排序的链表，请将它们合并成一个排序的链表。

分析：利用最小堆选取值最小的节点

用k个指针分别指向这k个链表的头节点，每次从这k个节点中选取值最小的节点。然后将指向值最小的节点的指针向后移动一步，再比较k个指针指向的节点并选取值最小的节点。重复这个过程，直到所有节点都被选取出来。

解

public class Test {public static void main(String[] args) {ListNode listNode1 = new ListNode(1);ListNode listNode2 = new ListNode(2);ListNode listNode3 = new ListNode(3);ListNode listNode4 = new ListNode(4);ListNode listNode5 = new ListNode(5);ListNode listNode6 = new ListNode(6);ListNode listNode7 = new ListNode(7);ListNode listNode8 = new ListNode(8);ListNode listNode9 = new ListNode(9);listNode1.next = listNode4;listNode4.next = listNode7;listNode2.next = listNode5;listNode5.next = listNode8;listNode3.next = listNode6;listNode6.next = listNode9;ListNode[] lists = {listNode1, listNode2, listNode3};ListNode result = mergeKLists(lists);while (result != null) {System.out.println(result.val);result = result.next;}}public static ListNode mergeKLists(ListNode[] lists) {ListNode dummy = new ListNode(0);ListNode cur = dummy;PriorityQueue<ListNode> minHeap = new PriorityQueue<>((n1, n2) -> n1.val - n2.val);for (ListNode list : lists) {if (list != null) {minHeap.offer(list);}}while (!minHeap.isEmpty()) {ListNode least = minHeap.poll();cur.next = least;cur = least;if (least.next != null) {minHeap.offer(least.next);}}return dummy.next;}
}

分析：按照归并排序的思路合并链表

输入的k个排序链表可以分成两部分，前k/2个链表和后k/2个链表。如果将前k/2个链表和后k/2个链表分别合并成两个排序的链表，再将两个排序的链表合并，那么所有链表都合并了。合并k/2个链表与合并k个链表是同一个问题，可以调用递归函数解决。

解

public class Test {public static void main(String[] args) {ListNode listNode1 = new ListNode(1);ListNode listNode2 = new ListNode(2);ListNode listNode3 = new ListNode(3);ListNode listNode4 = new ListNode(4);ListNode listNode5 = new ListNode(5);ListNode listNode6 = new ListNode(6);ListNode listNode7 = new ListNode(7);ListNode listNode8 = new ListNode(8);ListNode listNode9 = new ListNode(9);listNode1.next = listNode4;listNode4.next = listNode7;listNode2.next = listNode5;listNode5.next = listNode8;listNode3.next = listNode6;listNode6.next = listNode9;ListNode[] lists = {listNode1, listNode2, listNode3};ListNode result = mergeKLists(lists);while (result != null) {System.out.println(result.val);result = result.next;}}public static ListNode mergeKLists(ListNode[] lists) {if (lists.length == 0) {return null;}return mergeLists(lists, 0, lists.length);}private static ListNode mergeLists(ListNode[] lists, int start, int end) {if (start + 1 == end) {return lists[start];}int mid = (start + end) / 2;ListNode head1 = mergeLists(lists, start, mid);ListNode head2 = mergeLists(lists, mid, end);return merge(head1, head2);}private static ListNode merge(ListNode head1, ListNode head2) {ListNode dummy = new ListNode(0);ListNode cur = dummy;while (head1 != null && head2 != null) {if (head1.val < head2.val) {cur.next = head1;head1 = head1.next;}else {cur.next = head2;head2 = head2.next;}cur = cur.next;}cur.next = head1 == null ? head2 : head1;return dummy.next;}
}