刷题148. Sort List

一、题目说明

题目148. Sort List，对链表进行排序，时间复杂度要求是O(nlog(n))，空间复杂度要求是常量。难度是Medium！

二、我的解答

根据要求，唯一符合标准的是归并排序。

class Solution{
	public:
		ListNode* sortList(ListNode* head){
			if(head==NULL || head->next==NULL) return head;
			return mergeSort(head);
		}
		//归并排序 
		ListNode* mergeSort(ListNode* node){
			if(node==NULL || node->next==NULL) return node;
			ListNode* fast= node,*slow = node;
			ListNode* breakNode = node;//breakNode 指向l1的最后一个元素 
			//也可以采用先遍历一遍，统计链表节点的数量，然后归并排序
			//找到链表中间 
			while(fast && fast->next){
				fast = fast->next->next;
				breakNode = slow;
				slow = slow->next;
			}
			breakNode->next = NULL;
			ListNode* l1 = mergeSort(node);
			ListNode* l2 = mergeSort(slow);
			
			return merge(l1,l2);
		} 
		
		//合并两个链表 recursive
		ListNode* merge(ListNode* l1,ListNode* l2){
			if(l1 == NULL){
				return l2;
			}
			if(l2 == NULL){
				return l1;
			}
			if(l1->val < l2->val){
				l1->next = merge(l1->next,l2);
				return l1;
			}else{
				l2->next = merge(l2->next,l1);
				return l2;
			}
		} 
};

性能如下：

Runtime: 60 ms, faster than 41.32% of C++ online submissions for Sort List.
Memory Usage: 16.2 MB, less than 15.00% of C++ online submissions for Sort List.

三、优化措施

将merge函数，修改为非递归版本：

class Solution{
	public:
		ListNode* sortList(ListNode* head){
			if(head==NULL || head->next==NULL) return head;
			return mergeSort(head);
		}
		//归并排序 
		ListNode* mergeSort(ListNode* node){
			if(node==NULL || node->next==NULL) return node;
			ListNode* fast= node,*slow = node;
			ListNode* breakNode = node;//breakNode 指向l1的最后一个元素 
			//找到链表中间 
			while(fast && fast->next){
				fast = fast->next->next;
				breakNode = slow;
				slow = slow->next;
			}
			breakNode->next = NULL;
			ListNode* l1 = mergeSort(node);
			ListNode* l2 = mergeSort(slow);
			
			return merge(l1,l2);
		} 
		//合并两个链表
		ListNode* merge(ListNode* l1,ListNode* l2){
			if(l1 == NULL) return l2;
			if(l2 == NULL) return l1;
			ListNode dummy(0);
			ListNode* p = &dummy;
			while(l1!=NULL && l2!=NULL){
				if(l1->val < l2->val){
					p->next = l1;
					l1 = l1->next;	
				}else{
					p->next = l2;
					l2 = l2->next;
				}
				p = p->next;
			}
			if(l1 !=NULL){
				p->next = l1;
			}
			if(l2 !=NULL){
				p->next = l2;
			}
			return dummy.next;
		}
};

性能如下：

Runtime: 52 ms, faster than 67.44% of C++ online submissions for Sort List.
Memory Usage: 15 MB, less than 15.00% of C++ online submissions for Sort List.