CONY的世界

23. Merge k Sorted Lists

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

就像書上所說，如果結果不注重中間的排序，只在意最小或是最大，可以用heap

這邊就是可以視作先把各個listnode 塞進去，然後一個一個用 min 拉出來

因為是用到heap，time complexity是O(nlogk)，space complexity是O(k)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists==null||lists.length==0) return null;
        PriorityQueue<ListNode> queue = new PriorityQueue(lists.length, new Comparator<ListNode>(){
            public int compare(ListNode a, ListNode b){
                return a.val-b.val;
            }
        });
        
        for(ListNode list:lists) {
            if(list != null) 
               queue.add(list);
        }
        
        ListNode dummy = new ListNode(0), tail = dummy;
        while(!queue.isEmpty()){
            tail.next = queue.poll();
            tail = tail.next;
            if(tail.next != null)
                queue.add(tail.next);
        }
        return dummy.next;
    }
}