题解 | #合并k个已排序的链表#

/*
 * function ListNode(x){
 *   this.val = x;
 *   this.next = null;
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param lists ListNode类一维数组
 * @return ListNode类
 */
function mergeTwoLists(l1, l2) {
    if (!l1) return l2;
    if (!l2) return l1;
    if (l1.val < l2.val) {
        l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    } else {
        l2.next = mergeTwoLists(l2.next, l1);
        return l2;
    }
}
function merge(lists) {
    let middle = Math.floor(lists.length / 2);
    let left = mergeKLists(lists.slice(0, middle));
    let right = mergeKLists(lists.slice(middle));
    return mergeTwoLists(left, right);
}
// 以return 的形式 简化 多个if判断
function mergeKLists(lists) {
    return lists.length === 0
        ? null
        : lists.length === 1
        ? lists[0]
        : lists.length === 2
        ? mergeTwoLists(lists[0], lists[1])
        : merge(lists);
}
module.exports = {
    mergeKLists: mergeKLists,
};