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合并K个升序链表 ​

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
javascript
/**
 * 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： ​

javascript
/**
 * 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;
}