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。

示例 1:

输入:lists = [[1,4,5],[1,3,4],[2,6]]
输出:[1,1,2,3,4,4,5,6]
解释:链表数组如下:
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

示例 2:

输入:lists = []
输出:[]

示例 3:

输入:lists = [[]]
输出:[]

提示:

  • k == lists.length
  • 0 <= k <= 10^4
  • 0 <= lists[i].length <= 500
  • -10^4 <= lists[i][j] <= 10^4
  • lists[i] 按 升序 排列
  • lists[i].length 的总和不超过 10^4
/** * Definition for singly-linked list. * function ListNode(val, next) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } */ /** * @param {ListNode[]} lists * @return {ListNode} */ var mergeKLists = function(lists) { };

参考答案

/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode[]} lists * @return {ListNode} */ var mergeKLists = function(lists) { if (lists.length === 0) return null; return mergeArr(lists); }; function mergeArr(lists) { if (lists.length <= 1) return lists[0]; let index = Math.floor(lists.length / 2); const left = mergeArr(lists.slice(0, index)) const right = mergeArr(lists.slice(index)); return merge(left, right); } function merge(l1, l2) { if (l1 == null && l2 == null) return null; if (l1 != null && l2 == null) return l1; if (l1 == null && l2 != null) return l2; let newHead = null, head = null; while (l1 != null && l2 != null) { if (l1.val < l2.val) { if (!head) { newHead = l1; head = l1; } else { newHead.next = l1; newHead = newHead.next; } l1 = l1.next; } else { if (!head) { newHead = l2; head = l2; } else { newHead.next = l2; newHead = newHead.next; } l2 = l2.next; } } newHead.next = l1 ? l1 : l2; return head; }