剑指 Offer II 078. 合并排序链表
给定一个链表数组，每个链表都已经按升序排列。
请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.length0 <= k <= 10^40 <= lists[i].length <= 500-10^4 <= lists[i][j] <= 10^4lists[i] 按 升序 排列lists[i].length 的总和不超过 10^4
注意：本题与主站 23 题相同： https://leetcode.cn/problems/merge-k-sorted-lists/
原站题解
java 解法, 执行用时: 4 ms, 内存消耗: 43.2 MB, 提交时间: 2023-01-12 10:16:16
/**
* 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; }
* }
*/
// 利用优先队列合并
class Solution {
class Status implements Comparable<Status> {
int val;
ListNode ptr;
Status(int val, ListNode ptr) {
this.val = val;
this.ptr = ptr;
}
public int compareTo(Status status2) {
return this.val - status2.val;
}
}
PriorityQueue<Status> queue = new PriorityQueue<Status>();
public ListNode mergeKLists(ListNode[] lists) {
for (ListNode node: lists) {
if (node != null) {
queue.offer(new Status(node.val, node));
}
}
ListNode head = new ListNode(0);
ListNode tail = head;
while (!queue.isEmpty()) {
Status f = queue.poll();
tail.next = f.ptr;
tail = tail.next;
if (f.ptr.next != null) {
queue.offer(new Status(f.ptr.next.val, f.ptr.next));
}
}
return head.next;
}
}
java 解法, 执行用时: 1 ms, 内存消耗: 43.7 MB, 提交时间: 2023-01-12 10:15:50
/**
* 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; }
* }
*/
// 分治合并
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
return merge(lists, 0, lists.length - 1);
}
public ListNode merge(ListNode[] lists, int l, int r) {
if (l == r) {
return lists[l];
}
if (l > r) {
return null;
}
int mid = (l + r) >> 1;
return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
}
public ListNode mergeTwoLists(ListNode a, ListNode b) {
if (a == null || b == null) {
return a != null ? a : b;
}
ListNode head = new ListNode(0);
ListNode tail = head, aPtr = a, bPtr = b;
while (aPtr != null && bPtr != null) {
if (aPtr.val < bPtr.val) {
tail.next = aPtr;
aPtr = aPtr.next;
} else {
tail.next = bPtr;
bPtr = bPtr.next;
}
tail = tail.next;
}
tail.next = (aPtr != null ? aPtr : bPtr);
return head.next;
}
}
java 解法, 执行用时: 98 ms, 内存消耗: 43.7 MB, 提交时间: 2023-01-12 10:14:40
/**
* 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; }
* }
*/
// 顺序合并
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode ans = null;
for (int i = 0; i < lists.length; ++i) {
ans = mergeTwoLists(ans, lists[i]);
}
return ans;
}
public ListNode mergeTwoLists(ListNode a, ListNode b) {
if (a == null || b == null) {
return a != null ? a : b;
}
ListNode head = new ListNode(0);
ListNode tail = head, aPtr = a, bPtr = b;
while (aPtr != null && bPtr != null) {
if (aPtr.val < bPtr.val) {
tail.next = aPtr;
aPtr = aPtr.next;
} else {
tail.next = bPtr;
bPtr = bPtr.next;
}
tail = tail.next;
}
tail.next = (aPtr != null ? aPtr : bPtr);
return head.next;
}
}
python3 解法, 执行用时: 68 ms, 内存消耗: 18.6 MB, 提交时间: 2023-01-12 10:12:40
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
if not lists:return
import heapq
queue=[]
for l in lists:#将lists的值放进小根堆
head=l
while head:
heapq.heappush(queue,head.val)
head=head.next
dummy=ListNode(0)#构造虚拟节点
cur=dummy
while queue:#将堆顶取出连接成链表
cur.next=ListNode(heapq.heappop(queue))
cur=cur.next
return dummy.next