27 June 2014

Merge k sorted Lists

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

解法:这题有两种较好的解法,一种解法是类似锦标赛,每次将链表集合一分为2,递归求解。第二种解法是每次取出k个链表的链表头,将他们放入堆中,每次从取出堆中取出最小元素,然后将最小元素的下一个运算插入堆,再维护,直到堆空,排序完成。两种解法的复杂度都是O(nklogk),n为链表最大长度。第一次提交时是每次取出一个链表和长链表合并,结果超时,锦标赛顺利过了。

代码如下:

//{java}
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        if(lists.size()==0)
            return null;
        if(lists.size()==1)
            return lists.get(0);
        List<ListNode> aa = lists.subList(0,lists.size()/2);
        List<ListNode> bb = lists.subList(lists.size()/2,lists.size());
        ListNode a = mergeKLists(aa);
        ListNode b = mergeKLists(bb);
        ListNode head = new ListNode(0);
        ListNode p = head;
        while(a!=null && b!=null)
        {
            if(a.val>b.val)
            {
                p.next=b;
                b=b.next;
            }else
            {
                p.next=a;
                a=a.next;
            }
            p=p.next;
        }
        if(a!=null)
            p.next=a;
        if(b!=null)
            p.next=b;
        return head.next;
    }
}


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