标签：dup 代码 ret tput span time lex slow 思路
算法描述：
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：时间复杂度O(nlogn)。写了快排感觉很奇怪，看了其他人答案，发现都是归并排序。
快排代码很乱，如下。
ListNode* sortList(ListNode* head) { if(head==nullptr || head->next==nullptr) return head; ListNode* leftHead = new ListNode(INT_MIN); ListNode* rightHead = new ListNode(INT_MIN); ListNode* lc = leftHead; ListNode* rc = rightHead; ListNode* pivot = head; ListNode* cur = head->next; while(cur!=nullptr){ if(cur->val < pivot->val){ lc->next = cur; lc = lc->next; }else { rc->next = cur; rc = rc->next; } cur=cur->next; } lc->next = nullptr; rc->next = nullptr; leftHead->next = sortList(leftHead->next); rightHead->next = sortList(rightHead->next); cur = leftHead; while(cur->next!=nullptr) cur=cur->next; cur->next=pivot; cur->next->next = rightHead->next; return leftHead->next; }
归并排序代码：
ListNode* sortList(ListNode* head) { if(head==nullptr || head->next==nullptr) return head; ListNode* fast = head->next; ListNode* slow = head; while(fast!=nullptr && fast->next!=nullptr){ fast=fast->next->next; slow=slow->next; } fast = slow->next; slow->next = nullptr; return merge(sortList(fast), sortList(head)); } ListNode* merge(ListNode* l1, ListNode* l2){ ListNode* dup = new ListNode(-1); ListNode* cur = dup; while(l1!=nullptr && l2!=nullptr){ if(l1->val < l2->val){ cur->next=l1; l1=l1->next; }else{ cur->next=l2; l2=l2->next; } cur=cur->next; } if(l1!=nullptr) cur->next = l1; else cur->next = l2; return dup->next; }
标签：dup 代码 ret tput span time lex slow 思路
原文地址：https://www.cnblogs.com/nobodywang/p/10354102.html
