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