leetcode题解-23. Merge k Sorted Lists

题意：合并k个有序的链表，分析描述算法的复杂度。

分析：这道题目在分布式系统中非常常见，来自不同client的sorted list要在central server上面merge起来。解法就是有点类似于归并排序的思路，就是分治法，不了解归并排序的朋友，请参见归并排序-维基百科，是一个比较经典的O(nlogn)的排序算法，还是比较重要的。

思路是先分成两个子任务，然后递归求子任务，最后回溯回来。这个题目也是这样，先把k个list分成两半，然后继续划分，知道剩下两个list就合并起来，合并时需要用到两个有序的链表子过程。不熟悉的朋友可以复习一下。代码如下：

class Solution {    public static ListNode mergeKLists(ListNode[] lists) {        if(lists.length == 0) return null;        return help(lists, 0, lists.length - 1);    }    public static ListNode help(ListNode[] lists, int l, int r) {        int m = (l + r) / 2;        if(l < r){            return mergeTwoLists(help(lists, l, m), help(lists, m + 1, r));        }        return lists[l];    }    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {        if(l1 == null || l2 == null) return l1==null? l2:l1;        ListNode dummy = new ListNode(0);        if(l1.val < l2.val){            dummy.next = l1;            l1 = l1.next;        }else{            dummy.next = l2;            l2 = l2.next;        }        ListNode temp = dummy.next;        while(l1 != null  && l2 != null){            if(l1.val < l2.val){                temp.next = l1;                temp = temp.next;                l1 = l1.next;            }else{                temp.next = l2;                temp = temp.next;                l2 = l2.next;            }        }        temp.next = l1 == null ? l2:l1;        return dummy.next;    }    public static void print(ListNode head){        while(head != null){            System.out.println(head.val);            head = head.next;        }    }    public static void main(String[] args) {        ListNode l1 = new ListNode(4);        ListNode l2 = new ListNode(5);        ListNode l3 = new ListNode(6);        ListNode l4 = new ListNode(1);        ListNode l5 = new ListNode(2);        ListNode l6 = new ListNode(3);        l1.next = l2;        l2.next = l3;        l4.next = l5;        l5.next = l6;        ListNode[] lists = {l1,l4};        print(mergeKLists(lists));    }}