描述

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

样例

Given lists:
[
  2->4->null,
  null,
  -1->null
],
return -1->2->4->null.

代码实现

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param lists: a list of ListNode
     * @return: The head of one sorted list.
     */
   
    public ListNode mergeKLists(List<ListNode> lists) {  
        if (lists.size() == 0 || lists == null) {
            return null;
        }
        return mergeHelper(lists, 0, lists.size()-1);
    }
     // divde and conquer
    private ListNode mergeHelper(List<ListNode> lists, int start, int end) {
        if (start == end) {
            return lists.get(start);
        }
        
        int mid = start + (end - start) / 2;
        ListNode left = mergeHelper(lists, start, mid);
        ListNode right = mergeHelper(lists, mid+1, end);
        return mergeTwoLists(left, right);
    }
    //merge two sorted lists
    private ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while (list1 != null && list2 != null) {
            if (list1.val > list2.val) {
                tail.next = list2;
                tail = tail.next;
                list2 = list2.next;
            } else {
                tail.next = list1;
                tail = tail.next;
                list1 = list1.next;
            }
        }
        if (list1 != null) {
            tail.next = list1;
        } else {
            tail.next = list2;
        }
        return dummy.next;
    }
}