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

方法1：(利用類似的二路歸併排序，遞迴的方式)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
// 利用二路歸併排序的思想
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
    if(lists == null || lists.length < 1) {
            return null;
        }
        return helper(lists, 0, lists.length-1);
    }
    // 終止條件是l和r相等，最主要就是這個迴圈控制語句
    private ListNode helper(ListNode[] lists, int l, int r) {
        if(l == r) {
            return lists[l];
        }
        int mid = (l + r) / 2;
        return merge(helper(lists, l, mid), helper(lists, mid+1, r));
    }
    // 通過指標合併兩個連結串列
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while(l1 != null && l2 != null) {
            if(l1.val < l2.val) {
                tail.next = l1;
                // tail.next=new ListNode(l1.val);
                l1 = l1.next;
            } else {
                tail.next = l2;
                // tail.next=new ListNode(l2.val);
                l2 = l2.next;
            }
            tail = tail.next;
        }
        // 如果l1不為空的話插入到tail後面
        if(l1 != null) {
            tail.next = l1;
        }
        // 如果l2不為空的話插入tail的後面
        if(l2 != null) {
            tail.next = l2;
        }
        return dummy.next;
    }
}
 

時間複雜度：O(logn.n)

空間複雜度：O(n)