js 合并K个升序链表

leetcode 23.合并K个升序链表
采用以下循环的方式，运行时间过长，因此可以采取第二种方式：递归函数调用栈的方式

/**
 * 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) {
    if (lists.length == 0) return null;
    return mergeArr(lists);
};

// 采取循环的方式进行合并
function mergeArr(lists) {
    if(lists.length == 1) return lists[0];
    // 大于1
    let head = merge(lists[0],lists[1]);
    for (let i = 2; i < lists.length; i++) {
        head = merge(head,lists[i]);
    }
    return head;
}

// 两个两个进行合并
function merge(lis1,lis2) {
    if(lis1 == null && lis2 == null) return null;
    if(lis1 == null && lis2 != null) return lis2;
    if(lis1 != null && lis2 == null) return lis1;
    let lis = null,head = null;
    while(lis1 != null && lis2 != null) {
        if (lis1.val < lis2.val) {
            // 还没有头部
            if(!head) {
                lis = lis1;
                head = lis1;
            } else {
                lis.next = lis1;
                lis = lis.next;
            }
            lis1 = lis.next;
        } else {
            if (!head) {
                lis = lis2;
                head = lis2;
            } else {
                lis.next = lis2;
                lis = lis.next;
            }
        }
    }
    // 查看是否还有多余的数据
    lis.next = lis1 ? lis2 : lis1;
    return head;
}

/**
 * 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;
}