首先我想到的是采用一般递归法,将K个链表合并化为(k-1)两个链表合并 class Solution: def mergeKLists(self, lists: List[ListNode]) -> ListNode: # 运用递归, 化K个链表相加为两个链表相加 if not lists : return n = len(lists) for i in range(1,n): lists[0] = self.mergeList(lists[0],lists[i]) return lists[0…
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Example: Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6 合并 k 个排序链表,返回合并后的排序链表.请分析和描述算法的复杂度. 示例:…