23. Merge k Sorted Lists

23. Merge k Sorted Lists

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

var mergeKLists = function(lists) {
    
    var tempArry = [], listsLength;
    
    //先剔除空列表
    for (var i = 0; i < lists.length; ) {
        if (lists[i]) {
            i++;
        } else {
            lists.splice(i, 1);
        }
    }
    if (lists.length === 0) return [];
    
   //用一个队列来模拟，可以想象BFS的递推写法。
    while (lists.length > 1) {
        listsLength = lists.length;
        if (listsLength % 2 !== 0) {
            lists.push(null);
            listsLength++;
        }
        for (var i = 0; i < listsLength; i += 2) {
            tempArry.push(merge2Lists(lists[i], lists[i + 1]));
        }
        lists = tempArry;
        tempArry = [];
    }
    return lists[0];
    
    function merge2Lists(list1, list2) {
        var head = new ListNode(0), temp = head;
        while (list1 && list2) {
            if (list1.val < list2.val) {
                temp.next = new ListNode(list1.val);
                list1 = list1.next;
            } else {
                temp.next = new ListNode(list2.val);
                list2 = list2.next;
            }
            temp = temp.next;
        }
        temp.next = list1 ? list1 : list2;
        return head.next;
    }
};