标签：constant 快速排序 使用 inpu exit link complex ast const
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
Solution:
复杂度为O(nlogn)有快速排序，并归排序，堆排序，对于来列表而言，堆排序最适合了
使用快慢指针将链表分为两部分
merge01为简洁版
1 class Solution { 2 public: 3 ListNode* sortList(ListNode* head) { 4 if (head == nullptr || head->next == nullptr)return head; 5 ListNode *slow = head, *fast = head, *pre = head; 6 while (fast != nullptr && fast->next != nullptr) 7 { 8 pre = slow; 9 slow = slow->next; 10 fast = fast->next->next; 11 } 12 pre->next = nullptr;//将链表分为两段 13 return merge(sortList(head), sortList(slow)); 14 } 15 ListNode* merge01(ListNode* l1, ListNode* l2) 16 { 17 if (l1 == nullptr || l2 == nullptr)return l1 == nullptr ? l2 : l1; 18 if (l1->val < l2->val) 19 { 20 l1->next = merge(l1->next, l2); 21 return l1; 22 } 23 else 24 { 25 l2->next = merge(l1, l2->next); 26 return l2; 27 } 28 } 29 ListNode* merge02(ListNode* l1, ListNode* l2) 30 { 31 ListNode* ptr = new ListNode(-1); 32 ListNode* p = ptr; 33 while (l1 != nullptr && l2 != nullptr) 34 { 35 if (l1->val < l2->val) 36 { 37 p->next = l1; 38 l1 = l1->next; 39 } 40 else 41 { 42 p->next = l2; 43 l2 = l2->next; 44 } 45 p = p->next; 46 } 47 if (l1 == nullptr)p->next = l2; 48 if (l2 == nullptr)p->next = l1; 49 return ptr->next; 50 } 51 };
标签：constant 快速排序 使用 inpu exit link complex ast const
原文地址：https://www.cnblogs.com/zzw1024/p/11768301.html
