/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
if(lists.size() == 0)
return null;
else if(lists.size() == 1)
return lists.get(0);
else if(lists.size() == 2)
return mergeTwoLists(lists.get(0),lists.get(1));
else
return mergeTwoLists(mergeKLists(lists.subList(0, lists.size()/2)),mergeKLists(lists.subList(lists.size()/2, lists.size())));
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null)
return l2;
else if (l2 == null)
return l1;
else if (l1 == null || l2 == null)
return null;
ListNode ln = new ListNode(0);
ListNode temp = ln;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
temp.next = l1;
l1 = l1.next;
} else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
if (l2 != null)
temp.next = l2;
else if (l1 != null)
temp.next = l1;
return ln.next;
}
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
if(lists.size() == 0)
return null;
else if(lists.size() == 1)
return lists.get(0);
else if(lists.size() == 2)
return mergeTwoLists(lists.get(0),lists.get(1));
else
return mergeTwoLists(mergeKLists(lists.subList(0, lists.size()/2)),mergeKLists(lists.subList(lists.size()/2, lists.size())));
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null)
return l2;
else if (l2 == null)
return l1;
else if (l1 == null || l2 == null)
return null;
ListNode ln = new ListNode(0);
ListNode temp = ln;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
temp.next = l1;
l1 = l1.next;
} else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
if (l2 != null)
temp.next = l2;
else if (l1 != null)
temp.next = l1;
return ln.next;
}
}
-------------------------------------------------
HINTS:
1. 合并多个List, 细化下来,就是合并2个。。。依次叠加。为了提高效率,利用二叉搜索树的思想合并 ====> 递归
2. 将list 一一等分切割,就有了===> list.sublist(startIndex, toIndex)
====================
Merge k sorted
linked lists and return it as one sorted list. Analyze and describe its complexity.