23. Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

我们采用分治的方法来解决这个问题，其有K个链表，不断将其划分（partition），再将其归并（merge）。

划分的部分并不难，将其不断分成两部分，但是需要注意的是可能出现start和end相等的情况，这时候就直接return lists[start]就可以了。

mid = (start + end) / 2
start -- mid                    （1）
mid + 1 -- end                  （2）

上面的（1）和（2）就不断的替换更新，不断作为参数写到partition函数内。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return partition(lists, 0, lists.length - 1);
    }

    public ListNode partition(ListNode[] lists, int start, int end) {
        if (start == end) {
            return lists[start];
        }

        if (start < end) {
            int mid = (start + end) / 2;
            ListNode l1 = partition(lists, start, mid);
            ListNode l2 = partition(lists, mid + 1, end);
            return mergeTwoLists(l1, l2);
        }

        return null;
    }

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l2 == null) return l1;
        if (l1 == null) return l2;

        if (l1.val < l2.val) {
            ListNode tmp = l1;
            tmp.next = mergeTwoLists(l1.next, l2);
            return tmp;
        } else {
            ListNode tmp = l2;
            tmp.next = mergeTwoLists(l1, l2.next);
            return tmp;
        }
    }
}

Java2

/**
 * 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.length == 0) return null;
        int interval = 1;
        while (interval < lists.length) {
            for (int i = 0; i + interval < lists.length; i = i + interval*2) {
                lists[i] = mergeTwo(lists[i], lists[i+interval]);
            }
            interval *= 2;
        }
        return lists[0];
    }
    
    ListNode mergeTwo(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode temp = result;
        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;
            }
        }
        if (l1 != null) temp.next = l1;
        if (l2 != null) temp.next = l2;
        return result.next;
    }
}