題目

合并k個排序鏈表，并且返回合并后的排序鏈表。嘗試分析和描述其復雜度。

樣例
給出3個排序鏈表[2->4->null,null,-1->null]，返回 -1->2->4->null

分析

按照前面實現(xiàn)的合并兩個排序的鏈表的方法，兩兩合并，如果是奇數(shù)個，最后記得再合并最后一個即可。

代碼

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param lists: a list of ListNode
     * @return: The head of one sorted list.
     */
    public ListNode mergeKLists(List<ListNode> lists) {  
        // write your code here
        if (lists == null || lists.size() == 0) {
            return null;
        }
        
        while(lists.size() > 1) {
            List<ListNode> newLists = new ArrayList<ListNode>();
            for(int i=0;i+1<lists.size();i=i+2) {
                ListNode mergedList = mergeTwoLists(lists.get(i), lists.get(i+1));
                newLists.add(mergedList);
            }
            
            if(lists.size()%2 == 1) {
                newLists.add(lists.get(lists.size()-1));
            }
            
            lists = newLists;
        }
        
        return lists.get(0);
        
    }
    
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode point = head;
        
        while(l1!=null && l2!=null) {
            if(l1.val<l2.val) {
                point.next = l1;
                l1 = l1.next;
            }
            else {
                point.next = l2;
                l2 = l2.next;
            }
            
            point = point.next;
        }
        
        if(l1!=null) {
            point.next = l1;
        }
        
        if(l2!=null) {
            point.next = l2;
        }
        
        return head.next;   
    }
}