LeetCode:Sort List

LeetCode:Sort List

Sort a linked list in  O ( n  log  n ) time using constant space complexity.

此题用归并排序。

第一次对单链表使用归并，对归并一直不是很熟悉，就此详细记录求解过程。

//此时head1和head2都是已排序的，将head2并入head1中
ListNode* merge(ListNode* head1, ListNode* head2)
{
	//新生成一个节点作为头结点，便于对head1头结点进行插入
	ListNode* head = new ListNode(0);
	head->next = head1;
	//pre记录插入位置前一节点
	ListNode* pre = head;

	while(head1!=NULL && head2!=NULL)
	{
		if(head1->val < head2->val)
		{
			head1 = head1->next;
		}
		else
		{
			//将节点插入head1中
			ListNode* tmp = head2->next;
			head2->next = pre->next;
			pre->next = head2;
			head2 = tmp;
		}
		//位置后移
		pre = pre->next;
	}

	//若head2还有，则接入head1的最末尾
	if(head2 != NULL)
	{
		pre->next = head2;
	}

	return head->next;
}

ListNode* mergeSort(ListNode* head)
{
	if(head==NULL || head->next==NULL)
	{
		return head;
	}

	ListNode* slow = head;
	ListNode* fast = head;

	//从head指向的链表中进行查找，让slow指向中点，fast指向末尾
	while(fast->next!=NULL && fast->next->next!=NULL)
	{
		slow = slow->next;
		fast = fast->next->next;
	}

	//从链表中间划分，head2指向后半部分
	ListNode* head2 = slow->next;
	//head1指向前半部分
	slow->next = NULL;
	ListNode* head1 = head;

	//进一步细分链表
	head1 = mergeSort(head1);
	head2 = mergeSort(head2);

	//合并两段
	return merge(head1, head2);
}

ListNode *sortList(ListNode *head) 
{
	return mergeSort(head);
}