Leetcode 23 Merge k Sorted Lists

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

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

這個題目為合併連結串列的升級版，可以使用優先順序佇列來實現。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        if(lists==null) return null;
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        ListNode dummy = new ListNode(0);
        ListNode nex = dummy;
        for(int i=0;i<lists.size();i++){
            ListNode cur = lists.get(i);
            while(cur!=null){
                pq.add(cur.val);
                cur = cur.next;               
            }
        }
        int n = pq.size();
        for(int i=0;i<n;i++){
            int t = pq.poll();
            nex.next = new ListNode(t);
            nex = nex.next;
        }
        return dummy.next;
    }
}

先新增到優先順序佇列中然後一個一個的取出來進行連結串列的新增

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public static ListNode mergeKLists(ListNode[] lists){
    return partion(lists,0,lists.length-1);
}

public static ListNode partion(ListNode[] lists,int s,int e){
    if(s==e)  return lists[s];
    if(s<e){
        int q=(s+e)/2;
        ListNode l1=partion(lists,s,q);
        ListNode l2=partion(lists,q+1,e);
        return merge(l1,l2);
    }else
        return null;
}

//This function is from Merge Two Sorted Lists.
public static ListNode merge(ListNode l1,ListNode l2){
    if(l1==null) return l2;
    if(l2==null) return l1;
    if(l1.val<l2.val){
        l1.next=merge(l1.next,l2);
        return l1;
    }else{
        l2.next=merge(l1,l2.next);
        return l2;
    }
}
}

不斷地進行遞迴得出結果