} ListNode*sortList(ListNode* head){if(!head)returnNULL;if(!head->next)returnhead; ListNode* res = new ListNode(0); ListNode* pos = res; ListNode* mid = getMid(head); ListNode* left = sortList(head); ListNode* right = sortList(mid);while(left&&right) {if(left->val < right->va...
在LeetCode 里面,因为只有归并排序的时间复杂度为O(1),所以快速排序用不了,前面两个都没用直接看最后一个归并排序。 冒泡排序(超时了) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 publicListNode sortList(ListNode head) { if(null== head) retur...
3、递归版 publicclassSolution{publicListNodesortList(ListNode head){// 1、递归结束条件if(head==null||head.next==null){returnhead;}// 2、找到链表中间节点并断开链表 & 递归下探ListNode midNode=middleNode(head);ListNode rightHead=midNode.next;midNode.next=null;// 分治,递归排序左右两个链表ListNo...
// 排序链表(LeetCode 148):https://leetcode.cn/problems/sort-list/ classSolution{ publicListNodesortList(ListNode head){ // 获链表的总长度 intlength =0; // 从链表的头节点开始访问 ListNode node = head; // 利用 while 循环,可以统计出链表的节点个数,即长度 while(node !=null) { length++;...
s=self.sortList(s) l=self.sortList(l) # 合并 3 个链表, 这一步复杂度是 o(n) # # 可以同时返回链表的头指针和尾指针加速链表的合并。 dummy=ListNode(0) cur=dummy# cur: 非 None 的尾节点 # p: 下一个需要连接的节点 forpin[s,e,l]: ...
def sortList(self, head: ListNode) -> ListNode: if not head or not head.next: return head # termination. # cut the LinkedList at the mid index. slow, fast = head, head.next while fast and fast.next: fast, slow = fast.next.next, slow.next ...
right = self.sortList(mid) return self.__merge(left, right) def __findMidNode(self, head): # 找到中间节点,断开链表,之后再__merge中合并两个链表 slow = fast = head while fast.next and fast.next.next: slow = slow.next fast = fast.next.next ...
Sort a linked list in O(n log n) time using constant space complexity. Example 1: Input: 4->2->1->3 Output: 1->2->3->4 Example 2: Input: -1->5->3->4->0 Output: -1->0->3->4->5 解题思路:回想排序算法,线性对数时间复杂度的有:快速排序,堆排序,归并排序,前两种暂时没实现...
class Solution { public: ListNode* sortList(ListNode* head) { if(head == NULL) return head; Quick_Sort(head, NULL); return head; } void Quick_Sort(ListNode* head, ListNode* end) { if (head != end && head->next != end) { ListNode* ptr = head; while(ptr->next != end) ptr...
class Solution { public: ListNode* sortList(ListNode* head) { if(head == NULL || head->next == NULL) return head; ListNode *fast = head->next, *slow = head, *rightHead; //fast初始化先走一步,偶数个链表时,防止一边0个,一边2个节点 while(fast && fast->next) { fast = fast->next...