[Leetcode] Merge Two Sorted Lists Merge K Sorted Lists 归并有序列表

Merge Two Sorted Lists

最新更新请见：https://yanjia.me/zh/2019/01/...

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

依次拼接

复杂度

时间 O(N) 空间 O(1)

思路

该题就是简单的把两个链表的节点拼接起来，我们可以用一个Dummy头，将比较过后的节点接在这个Dummy头之后。最后如果有没比较完的，说明另一个list的值全比这个list剩下的小，而且拼完了，所以可以把剩下的直接全部接上去。

代码

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        // 创建一个dummy头，从后面开始接
        ListNode dummy = new ListNode(0);
        ListNode curr = dummy;
        // 依次比较拼接
        while(l1 != null && l2 != null){
            if(l1.val <= l2.val){
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }
            curr = curr.next;
        }
        // 把剩余的全拼上去
        if(l1 == null){
            curr.next = l2;
        } else if (l2 == null){
            curr.next = l1;
        }
        return dummy.next;
    }
}

Merge k Sorted Lists

最新解法请见：https://yanjia.me/zh/2019/01/...

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

优先队列

复杂度

时间 O(NlogK) 空间 O(K)

思路

当我们归并k个列表时，最简单的方法就是，对于每次插入，我们遍历这K个列表的最前面的元素，找出K个中最小的再加入到结果中。不过如果我们用一个优先队列（堆），将这K个元素加入再找堆顶元素，每次插入只要logK的复杂度。当拿出堆顶元素后，我们再将它所在链表的下一个元素拿出来，放到堆中。这样直到所有链表都被拿完，归并也就完成了。

注意

因为堆中是链表节点，我们在初始化堆时还要新建一个Comparator的类。

代码

Java

public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists.length == 0) return null;
        ListNode dummy = new ListNode(0);
        PriorityQueue<ListNode> q = new PriorityQueue<ListNode>(11, new Comparator<ListNode>(){
            public int compare(ListNode n1, ListNode n2){
                return n1.val - n2.val;
            }
        });
        // 初始化大小为k的堆
        for(int i = 0; i < lists.length; i++){
            if(lists[i] != null) q.offer(lists[i]);
        }
        ListNode curr = dummy;
        while(!q.isEmpty()){
            // 拿出堆顶元素
            curr.next = q.poll();
            curr = curr.next;
            // 将堆顶元素的下一个加入堆中
            if(curr.next != null){
                q.offer(curr.next);    
            }
        }
        return dummy.next;
    }
}

Python

class HeapItem:
    def __init__(self, node):
        self.node = node
        self.val = node.val
    
    def __lt__(self, other):
        return self.val < other.val

class Solution:
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        heap = []
        for node in lists:
            if node is not None:
                heap.append(HeapItem(node))
        heapify(heap)
        dummy = ListNode(0)
        head = dummy
        while len(heap) != 0:
            item = heappop(heap)
            node = item.node
            head.next = node
            head = head.next
            if node.next is not None:
                heappush(heap, HeapItem(node.next))
        return dummy.next