leetcode 23. Merge k Sorted Lists-蒲公英云

一、问题分析

二、代码实现

1、将第一个链表之外的所有链表分别与第一个链表合并

2、采用类似归并排序迭代版本自底向上的写法

3、小顶堆

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

将k个有序链表合并成一个有序链表后返回

一、问题分析

测试用例：

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

二、代码实现

1、将第一个链表之外的所有链表分别与第一个链表合并

/**
 * 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;
        }
        ListNode first = lists[0];
        for (int i = 1; i<lists.length; i++) {
            first = mergeTwoLists(first, lists[i]);
        }
        return first;
    }
    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode tempHead = new ListNode(-1);
        ListNode tempTail = tempHead;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                tempTail.next = l1;
                l1 = l1.next;
                tempTail = tempTail.next;
            } else {
                tempTail.next = l2;
                l2 = l2.next;
                tempTail = tempTail.next;
            }
        }
        if (l1 != null) {
            tempTail.next = l1;
        }
        if (l2 != null) {
            tempTail.next = l2;
        }
        // return tempHead.next;
        ListNode temp = tempHead;
        tempHead = tempHead.next;
        temp.next = null;
        return tempHead;
    }
}

2、采用类似归并排序迭代版本自底向上的写法

/**
 * 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;
        }
        int step = 1;        //2*step的下标跨度作为一个小组，将小组的两个链表合并
        while (step < lists.length) {
            for (int i=0; i<lists.length; i += 2*step) {
                if (i+step >= lists.length) {
                    // mergeTwoLists(lists[i], null);
                    // [[],[1]] 输出[]
                    lists[i] = mergeTwoLists(lists[i], null);
                } else {
                    // mergeTwoLists(lists[i], lists[i+step]);
                    // [[],[1]]  输出[]
                    lists[i] = mergeTwoLists(lists[i], lists[i+step]);
                }
                // step = 2 * step;
                // [[],[-1,5,11],[],[6,10]]  输出[-1,5,11]
            }
            step = step * 2;
        }
        return lists[0];
    }
    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        //[[], [1]] 输出[]
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        ListNode tempHead = new ListNode(-1);
        ListNode tempTail = tempHead;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                tempTail.next = l1;
                l1 = l1.next;
                tempTail = tempTail.next;
            } else {
                tempTail.next = l2;
                l2 = l2.next;
                tempTail = tempTail.next;
            }
        }
        if (l1 != null) {
            tempTail.next = l1;
        }
        if (l2 != null) {
            tempTail.next = l2;
        }
        // return tempHead.next;
        ListNode temp = tempHead;
        tempHead = tempHead.next;
        temp.next = null;
        return tempHead;
    }
}

3、小顶堆

/**
 * 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;
        }
        //小顶堆
        Queue<ListNode> heap = new PriorityQueue(new Comparator<ListNode>(){
            @Override public int compare(ListNode l1, ListNode l2) { 
                return l1.val - l2.val; 
            }
        });
        //Queue<ListNode> heap = new PriorityQueue<>((l1, l2) -> l1.val - l2.val);
        //Queue<ListNode> heap = new PriorityQueue<>(lists.length, (l1,l2) -> Integer.compare(l1.val, l2.val));
        ListNode head = new ListNode(0), tail = head;
        for (ListNode node : lists) {
            if (node != null) {
                heap.offer(node);
            }
        }
        while (!heap.isEmpty()) {
            tail.next = heap.poll();
            tail = tail.next;
            if (tail.next != null) {
                heap.offer(tail.next);
            }
        }
        return head.next;
    }
}