leetcode -- Merge k Sorted Lists add code

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

[解题思路]

以前的解法的时间复杂度过高，通过在网上搜索，得到优化的时间复杂度：O(n*lgk)

维护一个大小为k的最小堆，每次得到一个最小值，重复n次

 /**
  * 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(ArrayList<ListNode> lists) {
         // Start typing your Java solution below
         // DO NOT write main() function
         ListNode head = null;
         int len = lists.size();
         if(len <= 1)
             return null;
         head = merge2List(lists.get(0), lists.get(1));
         for(int i = 2; i < len; i++){
             head = merge2List(lists.get(i), head);
         }
         return head;

     }

     public ListNode merge2List(ListNode node1, ListNode node2){
         ListNode head = null;
         ListNode tmp = head;
         while(node1 != null && node2 != null){
             if(node1.val <= node2.val){
                 ListNode node = new ListNode(node1.val);
                 tmp = node;
                 tmp = tmp.next;
             } else {
                 ListNode node = new ListNode(node2.val);
                 tmp = node;
                 tmp = tmp.next;
             }
             node1 = node1.next;
             node2 = node2.next;
         }

         while(node1 != null){
             ListNode node = new ListNode(node1.val);
             tmp = node;
             tmp = tmp.next;
             node1 = node1.next;
         }

         while(node2 != null){
             ListNode node = new ListNode(node2.val);
             tmp = node;
             tmp = tmp.next;
             node2 = node2.next;
         }
         return head;

     }
 }

上一版本有bug，修复如下：

 public class Solution {
     public ListNode mergeKLists(ArrayList<ListNode> lists) {
         // Start typing your Java solution below
         // DO NOT write main() function
         ListNode head = null;
         int len = lists.size();
         if(len == 0)
             return null;
         else if(len == 1){
             return lists.get(0);
         }
         head = merge2List(lists.get(0), lists.get(1));
         for(int i = 2; i < len; i++){
             head = merge2List(lists.get(i), head);
         }
         return head;

     }

     public ListNode merge2List(ListNode node1, ListNode node2){
         ListNode head = new ListNode(Integer.MIN_VALUE);
         ListNode tmp = head;
         while(node1 != null && node2 != null){
             if(node1.val <= node2.val){
                 ListNode node = new ListNode(node1.val);
                 tmp.next = node;
                 tmp = tmp.next;
                 node1 = node1.next;
             } else {
                 ListNode node = new ListNode(node2.val);
                 tmp.next = node;
                 tmp = tmp.next;
                 node2 = node2.next;
             }
         }

         if(node1 != null){
             tmp.next = node1;
         }

         if(node2 != null){
             tmp.next = node2;
         }
         head = head.next;
         return head;

     }
 }

http://blog.csdn.net/zyfo2/article/details/8682727

http://tech-wonderland.net/blog/leetcode-merge-k-sorted-lists.html