BM5 合并 k 个已排序的链表

解法一： 归并+递归，做这题之前先看BM4解发二比较好理解，时间复杂度nlog(n)，空间复杂度O(n)

import java.util.*;

public class Solution {
    public ListNode mergeKLists(ArrayList lists) {

        if (lists == null || lists.size() == 0) {
            return null;
        }
        return partion(lists, 0, lists.size() - 1);
    }

    private ListNode partion(ArrayList lists, int left, int right) {
        if (left >= right) {
            return lists.get(left);
        }
        int mid = (left + right) / 2;
        ListNode node1 = partion(lists, left, mid);
        ListNode node2 = partion(lists, mid + 1, right);
        return merge(node1, node2);
    }

    private ListNode merge(ListNode node1, ListNode node2) {
        if (node1 == null) {
            return node2;
        }
        if (node2 == null) {
            return node1;
        }
        ListNode pre = null;
        if (node1.val < node2.val) {
            pre = node1;
            pre.next = merge(node1.next, node2);
        } else {
            pre = node2;
            pre.next = merge(node1, node2.next);
        }
        return pre;
    }
}

方法二：归并+循环，时间复杂度O(nlog(n))，空间复杂度O(n)

import java.util.*;

public class Solution {
    public ListNode mergeKLists(ArrayList lists) {

        if (lists == null || lists.size() == 0) {
            return null;
        }
        return partion(lists, 0, lists.size() - 1);
    }

    private ListNode partion(ArrayList lists, int left, int right) {
        if (left >= right) {
            return lists.get(left);
        }
        int mid = (left + right) / 2;
        ListNode node1 = partion(lists, left, mid);
        ListNode node2 = partion(lists, mid + 1, right);
        return merge(node1, node2);
    }

    private ListNode merge(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }
        ListNode cur = new ListNode(-1);
        ListNode pre = cur;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                cur.next = list1;
                list1 = list1.next;
                cur = cur.next;
            } else if (list1.val > list2.val) {
                cur.next = list2;
                list2 = list2.next;
                cur = cur.next;
            } else {
                cur.next = list1;
                list1 = list1.next;
                cur = cur.next;

                cur.next = list2;
                list2 = list2.next;
                cur = cur.next;
            }
        }
        while (list1 != null) {
            cur.next = list1;
            list1 = list1.next;
            cur = cur.next;
        }
        while (list2 != null) {
            cur.next = list2;
            list2 = list2.next;
            cur = cur.next;
        }
        return pre.next;
    }
}