23. 合并K个排序链表

23. 合并K个排序链表

题目描述

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

实现1

每次取出两个链表，合并成一个

/**
 * 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 res = lists[0];

        for (int i = 1; i < lists.length; i++) {
            res = insert(res, lists[i]);
        }
        
        return res;
    }

    public ListNode insert(ListNode list, ListNode list2) {
        if(list == null) {
            return list2;
        }

        if(list2 == null) return list;

        if(list.val > list2.val) {
            ListNode temp = list;
            list = list2;
            list2 = temp;
        }

        ListNode pre = list;
        ListNode next = list.next;

        while(list2 != null && pre != null) {
            if(pre.val <= list2.val && next == null) {
                pre.next = list2;
                break;
            } else if(pre.val <= list2.val && next.val > list2.val) {
                pre.next = list2;
                list2 = list2.next;
                pre = pre.next;
                pre.next = next;
            } else {
                pre = next;
                next = next.next;
            }
        }

        return list;
    }
}

实现2

分治

/**
 * 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;
        }
        return merger(lists, 0, lists.length - 1);
    }
    
    public ListNode merger(ListNode[] lists, int left, int right) {
        if(left == right) {
            return lists[left];
        }
        int mid = left + (right - left) / 2;
        
        ListNode list1 = merger(lists, left, mid);
        ListNode list2 = merger(lists, mid + 1, right);
        
        return mergerTwoLists(list1, list2);
    }
    
    public ListNode mergerTwoLists(ListNode list1, ListNode list2) {
        if(list1 == null) return list2;
        if(list2 == null) return list1;
        
        if(list1.val > list2.val) {
            ListNode temp = list1;
            list1 = list2;
            list2 = temp;
        }
        
        ListNode pre = list1;
        ListNode next = list1.next;
        
        while(list2 != null) {
            if(pre.val <= list2.val && next == null) {
                pre.next = list2;
                break;
            } else if(pre.val <= list2.val && next.val > list2.val) {
                ListNode temp = list2;
                list2 = list2.next;
                temp.next = next;
                pre.next = temp;
                pre = temp;
            } else {
                pre = next;
                next = next.next;
            }
        }
        
        return list1;
    }
}

来源：CSDN

作者：csdnyq

链接：https://blog.csdn.net/csdnyq/article/details/103757201