LeetCode 23. Merge k Sorted Lists(java)

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

解法一：用heap来做，先把k个list的第一个元素放进heap，每次poll出一个后，放进poll出的那个的后一个，直到heap为空。时间复杂度O（mklogk），空间复杂度为O（k）.

public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        if (lists.length == 1) return lists[0];
        PriorityQueue queue = new PriorityQueue<>(new Comparator() {
            public int compare(ListNode o1, ListNode o2) {
                return o1.val - o2.val;
            }
        });
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) queue.add(lists[i]);
        }
        ListNode dummy = new ListNode(-1), cur = dummy;
        while (!queue.isEmpty()) {
            ListNode temp = queue.poll();
            cur.next = temp;
            cur = temp;
            if (temp.next != null) queue.add(temp.next);
        }
        return dummy.next;
    }

解法二：merge sort的想法，比解法一更优。每次都两两list merge，直到merge成一个list，时间复杂度O（(m/2)(k/2 + k/4 + k/8 + …)logk），还是O（kmlogk），但是比方法一有常数量上的优化。空间复杂度为O（1）.

public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        int begin = 0, end = lists.length - 1;
        while (begin < end) {
            int mid = (begin + end - 1) / 2;
            for (int i = 0; i <= mid; i++) {
                lists[i] = merge2list(lists[i], lists[end - i]);
            }
            end = (begin + end) / 2;
        }
        return lists[0];
    }
    public ListNode merge2list(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) return null;
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode dummy = new ListNode(-1), cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                cur = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                cur = l2;
                l2 = l2.next;
            }
        }
        if (l1 != null) cur.next = l1;
        if (l2 != null) cur.next = l2;
        return dummy.next;
    }