️ 力扣原文
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。
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
输入:lists = []
输出:[]
输入:lists = [[]]
输出:[]
使用优先队列合并:
将所有链表的节点值存到一个优先级队列里面,它会自动进行排序,最后再将排序后的每个元素构造成一个新的链表。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
PriorityQueue<Integer> queue = new PriorityQueue<Integer>((n1, n2) -> n1-n2);
for (ListNode node : lists) {
while (node!=null){
queue.add(node.val);
node = node.next;
}
}
ListNode dummy = new ListNode(-1);
ListNode curr = dummy;
while (queue.peek()!=null){
curr.next = new ListNode(queue.poll());
curr = curr.next;
}
return dummy.next;
}
}
分治法:
将链表数组分成两个部分,然后对每个部分递归调用 mergeKLists 函数,最终合并两个链表,得到最终合并后的链表。
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if(lists.length == 0){return null;}
return merge(lists, 0, lists.length - 1);
}
public ListNode merge(ListNode[] lists, int left, int right){
if(left == right){
return lists[left];
}
int mid = (left + right)/2;
ListNode l1 = merge(lists, left, mid);
ListNode l2 = merge(lists, mid+1, right);
return mergeTwoLists(l1, l2);
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2){
ListNode dummy = new ListNode(-1);
ListNode curr = dummy;
while (l1 != null && l2 != null){
if(l1.val < l2.val){
curr.next = l1;
l1 = l1.next;
}else {
curr.next = l2;
l2 = l2.next;
}
curr = curr.next;
}
if(l1 == null){
curr.next = l2;
}
if(l2 == null){
curr.next = l1;
}
return dummy.next;
}
}