LeetCode 23 Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

合并k个有序链表。

最直接的想法先合并两条链表，然后再与第三条链表进行合并，那么时间复杂度是o(2n+3n……kn)=o(nk²)

另一种想法是借鉴归并排序的思想，两两合并。T(n)=2T(n/c)+cn,可得复杂度为o(klogn)

还可以使用堆排序的思想。把每个链表的头结点放入堆中。每次找到最小的加入到新的链表中。建堆复杂度为klog(k)，每次处理新节点为log(k),总复杂度nklog(k)。

代码如下：

/**链表合并 * @param lists * @return */public static ListNode mergeKLists(ListNode[] lists) {if (lists.length == 0)return null;if (lists.length == 1)return lists[0];ListNode head = new ListNode(0);ListNode resultList = head;int heapSize = 0;for (ListNode node : lists)if (node != null)heapSize++;lists = initHeap(lists, heapSize);while (heapSize > 0) {ListNode node = new ListNode(lists[0].val);head.next = node;head = head.next;if (lists[0].next == null) {lists[0] = lists[heapSize - 1];heapSize--;lists = minHeap(lists, 0, heapSize);} else {lists[0] = lists[0].next;lists = minHeap(lists, 0, heapSize);for (ListNode n : lists)System.out.print(n.val + ",");System.out.println();}}return resultList.next;}/** 初始化堆 * @param lists * @param heapSize * @return */public static ListNode[] initHeap(ListNode[] lists, int heapSize) {for (int i = 0; i < lists.length; i++) {if (lists[i] == null) {for (int j = lists.length - 1; j > i; j--) {if (lists[j] != null) {lists[i] = lists[j];break;}}}}int l = (int) (Math.ceil(heapSize / 2) - 1);while (l >= 0) {lists = minHeap(lists, l, heapSize);l--;}return lists;}/**最小化堆 * @param lists * @param l * @param heapSize * @return */public static ListNode[] minHeap(ListNode[] lists, int l, int heapSize) {int min = 0;if (l * 2 + 1 > heapSize - 1) {return lists;} else if (l * 2 + 2 > heapSize - 1) {// only has right childmin = l * 2 + 1;} else {// has left and right childif (lists[l * 2 + 2].val > lists[l * 2 + 1].val)min = l * 2 + 1;elsemin = l * 2 + 2;}if (lists[min].val < lists[l].val) {ListNode node = lists[min];lists[min] = lists[l];lists[l] = node;if (2 * min + 1 <= heapSize - 1)lists = minHeap(lists, min, heapSize);} else {return lists;}return lists;}