LeetCode -- Merge k Sorted Lists (Divide and Conquer / PriorityQueue)

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

分支策略：每次归并两个已排好序的链表，直至只剩下一个链表。

public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        //处理特殊情况
        if (lists == null)
            return null;
        int length = lists.size();
        if (length == 0)
            return null;
        if (length == 1)
            return lists.get(0);
        //
        ListNode low, high;
        while (length != 1){//逐个归并有序串
            int index;
            for (index = 0; index < length/2; ++index){
                low = lists.get(2*index);
                high = lists.get(2*index+1);
                lists.set(index, merge2Lists(low, high));
            }
            if (length%2 == 1)
                lists.set(index, lists.get(length-1));
            length = (length + 1) / 2;
        }
        return lists.get(0);
    }
    public ListNode merge2Lists(ListNode low, ListNode high){
        ListNode result = new ListNode(-1);
        ListNode p = result;
        while (low != null && high != null){
            if (low.val < high.val){
                p.next = low;
                low = low.next;
                p = p.next;
            } else {
                p.next = high;
                high = high.next;
                p = p.next;
            }
        }
        if (low == null)
            p.next = high;
        else
            p.next = low;
        return result.next;
    }
}

优先队列(或者说堆排序)：http://zh.wikipedia.org/zh/%E5%84%AA%E5%85%88%E4%BD%87%E5%88%97

public class Solution {
//此段代码出自https://oj.leetcode.com/discuss/25518/13-lines-in-java
    public ListNode mergeKLists(List<ListNode> lists) {
        Queue<ListNode> heap = new PriorityQueue(new Comparator<ListNode>(){
            @Override public int compare(ListNode l1, ListNode l2) {
                return l1.val - l2.val;
            }
        });
        ListNode head = new ListNode(0), tail = head;
        for (ListNode node : lists) if (node != null) heap.offer(node);
        while (!heap.isEmpty()) {
            tail.next = heap.poll();
            tail = tail.next;
            if (tail.next != null) heap.offer(tail.next);
        }
        return head.next;
    }
}