lintcode-medium-Merge k Sorted Lists

最新推荐文章于 2022-05-09 00:28:10 发布

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最新推荐文章于 2022-05-09 00:28:10 发布

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原文链接：http://www.cnblogs.com/goblinengineer/p/5339822.html

本文介绍了一种使用优先队列合并K个已排序链表的方法，并提供了完整的Java实现代码。该方法首先将所有链表的头节点放入优先队列中，然后依次取出最小节点并将其加入结果链表，直至队列为空。

Merge k sorted linked lists and return it as one sorted list.

Analyze and describe its complexity.

Example

Given lists:

[
  2->4->null,
  null,
  -1->null
],

return -1->2->4->null.

/**
 * 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;
        
        PriorityQueue<ListNode> queue = new PriorityQueue<ListNode>(lists.size(), new Comparator<ListNode>(){
                public int compare(ListNode p1, ListNode p2){
                    return p1.val - p2.val;
                }
            });
        ListNode fakehead = new ListNode(0);
        ListNode p = fakehead;
        
        for(int i = 0; i < lists.size(); i++){
            if(lists.get(i) != null)
                queue.offer(lists.get(i));
        }
        
        while(!queue.isEmpty()){
            ListNode temp = queue.poll();
            
            p.next = temp;
            p = p.next;
            
            if(temp.next != null)
                queue.offer(temp.next);
        }
        
        return fakehead.next;
    }
}