[leetcode] Merge k Sorted Lists

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

https://oj.leetcode.com/problems/merge-k-sorted-lists/

思路1（naive）：1跟2合并，然后跟3合并，然后跟4合并，一直到k个合并完。

    复杂度：1,2合并访问2n个节点，12与3合并访问3n个节点，...，123~k-1与k合并访问kn个节点，总共遍历节点数目n(3+4+5+...+k)，O(nk^2)。

思路2：mergeSort的思想，K个链表先划分为合并两个k/2的链表，每个k/2的链表在划分为k/4的链表，一直到只剩一个或者两个链表。

    复杂度：T(k)=2T(k/2)+O(nk)，根据 主定理（算法导论上有讲），复杂度为O(nklogk)。

思路3：维护一个大小为k的堆。每次取堆顶元素放到结果中，并把该元素的后继（如果有）放入堆中。

    复杂度：每个元素读取一次nk，堆操作logk，所以复杂度为O(nklogk)。

 

思路2代码：

public ListNode mergeKLists(ArrayList<ListNode> lists) {
        if (lists == null)
            return null;
        return merge(lists, 0, lists.size() - 1);

    }

    private ListNode merge(ArrayList<ListNode> lists, int start, int end) {
        if (start == end)
            return lists.get(start);
        int mid = (start + end) / 2;
        ListNode one = merge(lists, start, mid);
        ListNode two = merge(lists, mid + 1, end);
        return mergeTwoLists(one, two);
    }

    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null)
            return l2;
        if (l2 == null)
            return l1;
        ListNode p1 = l1;
        ListNode p2 = l2;

        ListNode head = new ListNode(-1);
        head.next = p1.val <= p2.val ? p1 : p2;
        ListNode tail = head;

        while (p1 != null && p2 != null) {
            if (p1.val <= p2.val) {
                tail.next = p1;
                ListNode oldP1 = p1;
                p1 = p1.next;
                oldP1.next = null;
                tail = oldP1;
            } else {
                tail.next = p2;
                ListNode oldP2 = p2;
                p2 = p2.next;
                oldP2.next = null;
                tail = oldP2;
            }

        }
        if (p1 != null)
            tail.next = p1;
        if (p2 != null)
            tail.next = p2;

        return head.next;

    }

 

 

 

思路3代码（加测试代码）：

import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Random;

public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        if (lists == null)
            return null;
        PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>(20, new Comparator<ListNode>() {
            @Override
            public int compare(ListNode o1, ListNode o2) {
                return o1.val - o2.val;
            }
        });
        ListNode head = new ListNode(-1);
        ListNode cur = head;
        for (int i = 0; i < lists.size(); i++) {
            if (lists.get(i) != null)
                pq.add(lists.get(i));
        }

        while (!pq.isEmpty()) {
            ListNode out = pq.remove();
            cur.next = out;
            cur = cur.next;
            if (out.next != null)
                pq.add(out.next);
        }

        return head.next;
    }

    public static void main(String[] args) {
        ArrayList<ListNode> lists = new ArrayList<ListNode>();
        int k = 3;
        for (int i = 0; i < k; i++) {
            ListNode list = makeList(getRandomArray());
            lists.add(list);
            printList(list);
        }
        printList(new Solution().mergeKLists(lists));

    }

    private static int[] getRandomArray() {
        Random rand = new Random();
        int size = rand.nextInt(10);
        int[] res = new int[size];
        if (size == 0)
            return res;
        res[size - 1] = 100;
        for (int i = size - 2; i >= 0; i--) {
            res[i] = rand.nextInt(res[i + 1] + 1);
        }
        return res;
    }

    private static ListNode makeList(int[] a) {
        ListNode head = new ListNode(-1);
        ListNode p = head;
        for (int i = 0; i < a.length; i++) {
            p.next = new ListNode(a[i]);
            p = p.next;
        }
        return head.next;

    }

    private static void printList(ListNode head) {
        while (head != null) {
            System.out.print(head.val);
            if (head.next != null)
                System.out.print("->");
            head = head.next;
        }
        System.out.println();

    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }
}

 

第二遍记录：

 

归并法：注意cornor case，lists==null or lists.size()==0.

 

public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        if(lists==null||lists.size()==0)
            return null;
        return merge(lists,0,lists.size()-1);
    }
    
    private ListNode merge(List<ListNode> lists, int start, int end){
        if(start>=end){
            return lists.get(start);
        }
        
        int mid = (start+end)/2;
        ListNode one = merge(lists,start,mid);
        ListNode two = merge(lists,mid+1,end);
        return mergeTwoLists(one,two);
    }
    
    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1==null)
            return l2;
        if(l2==null)
            return l1;
        ListNode dummyHead = new ListNode(-1);
        ListNode p =dummyHead;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                p.next =l1;
                l1=l1.next;
            }
            else{
                p.next=l2;
                l2=l2.next;
            }
            p=p.next;
        }
        if(l1!=null)
            p.next=l1;
        if(l2!=null)
            p.next=l2;
        
        return dummyHead.next;
    }    
}

 

 

 

用PriorityQueue的方法，注意自定义Comparator。

public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        if(lists==null)
            return null;
        PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>(20, new Comparator<ListNode>(){
            
            @Override
            public int compare(ListNode a, ListNode b){
                return a.val-b.val;
            }
        });
        
        ListNode dummyHead = new ListNode(-1);
        for(int i=0;i<lists.size();i++){
            ListNode head = lists.get(i);
            if(head!=null)
                pq.add(head);
        }
        
        ListNode cur = dummyHead;
        while(!pq.isEmpty()){
            ListNode out = pq.remove();
            cur.next = out;
            cur = cur.next;
            if(out.next!=null)
                pq.add(out.next);
        }
    
        return dummyHead.next;
    }
    
    
}

 

 

 

参考：

http://blog.csdn.net/linhuanmars/article/details/19899259

http://www.cnblogs.com/TenosDoIt/p/3673188.html