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0023-合并K个升序链表
https://leetcode.cn/problems/merge-k-sorted-lists
给你一个链表数组，每个链表都已经按升序排列。 请你将所有链表合并到一个升序链表中，返回合并后的链表。
示例 1：
输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2：
输入：lists = []
输出：[]
示例 3：
输入：lists = [[]]
输出：[]
提示：
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4
思路：逐次合并
写一个Merge两个有序链表的函数，然后逐次调用数组
csharp
public class Solution {
private ListNode MergeTwoLists(ListNode head1, ListNode head2){
ListNode dummyHead = new ListNode();
ListNode cur = dummyHead;
while(head1 != null && head2 != null){
if(head1.val < head2.val){
cur.next = head1;
head1 = head1.next;
}
else{
cur.next = head2;
head2 = head2.next;
}
cur = cur.next;
}
if(head1 != null){
cur.next = head1;
}
if(head2 != null){
cur.next = head2;
}
return dummyHead.next;
}
public ListNode MergeKLists(ListNode[] lists) {
ListNode head=null;
if(lists.Length > 0){
head = lists[0];
for(int i=1; i<lists.Length; i++){
head = MergeTwoLists(head,lists[i]);
}
}
return head;
}
}
思路：二分合并
将上面的链表使用二分法合并，减少合并次数。
csharp
public class Solution {
private ListNode MergeTwoLists(ListNode head1, ListNode head2){
ListNode dummyHead = new ListNode();
ListNode cur = dummyHead;
while(head1 != null && head2 != null){
if(head1.val < head2.val){
cur.next = head1;
head1 = head1.next;
}
else{
cur.next = head2;
head2 = head2.next;
}
cur = cur.next;
}
if(head1 != null){
cur.next = head1;
}
if(head2 != null){
cur.next = head2;
}
return dummyHead.next;
}
private ListNode MergeKLists(ListNode[] lists, int left, int right){
if(right < left){
return null;
}
else if(left == right){
return lists[left];
}
else{
int mid = left + (right - left) / 2;
ListNode head1 = MergeKLists(lists,left,mid);
ListNode head2 = MergeKLists(lists,mid+1,right);
return MergeTwoLists(head1,head2);
}
}
public ListNode MergeKLists(ListNode[] lists) {
return MergeKLists(lists,0,lists.Length-1);
}
}