https://leetcode.com/problems/sort-list/

對於連結串列的排序，非常適合用歸併排序演算法。但本題還要求只用O(1)的記憶體空間；如果考慮歸併演算法的棧的話，實際上本題的解法使用的是O(n)的記憶體空間。所以本題解法的時間複雜度是O(n Log(n))，空間複雜度是O(n)。其實歸併排序也可以使用遍歷的方法實現，每次遍歷使用不同的步長即可，但是本人懶得修改了～～

C++程式碼如下：

class Solution {
public:
	ListNode * sortList(ListNode * head) {
		if(head == NULL || head->next == NULL)
			return head;

		ListNode * walker = head;
		ListNode * runner = head;
		while(runner->next!=NULL && runner->next->next!=NULL) {
			walker = walker->next;
			runner = runner->next->next;
		}

		ListNode * head2 = walker->next;
		walker->next = NULL;
		ListNode * head1 = head;
		head1 = sortList(head1);
		head2 = sortList(head2);

		return merge(head1, head2);
	}

	ListNode * merge(ListNode * head1, ListNode * head2) {
		ListNode * helper = new ListNode(0);
		helper->next = head1;
		ListNode * pre = helper;
		while(head1!=NULL && head2!=NULL) {
			if(head1->val<head2->val)
				head1 = head1->next;
			else {
				ListNode * next = head2->next;
				head2->next = pre->next;
				pre->next = head2;
				head2 = next;
			}

			pre = pre->next;
		}
		
		if(head2!=NULL)
			pre->next = head2;

		return helper->next;
	}
};
 

Python程式碼如下：

class Solution:
	def sortList(self, head):
		if head == None or head.next == None:
			return head

		walker = head
		runner = head
		while runner.next != None and runner.next.next != None:
			walker = walker.next
			runner = runner.next.next

		head2 = self.sortList(walker.next)
		walker.next = None
		head1 = self.sortList(head)

		return self.merge(head1, head2)

	def merge(self, head1, head2):
		if head1 == None:
			return head2

		if head2 == None:
			return head1

		helper = ListNode(0)
		helper.next = head1
		pre = helper
		while head1 != None and head2 != None:
			if head1.val < head2.val:
				head1 = head1.next
			else:
				next = head2.next
				head2.next = head1
				pre.next = head2
				head2 = next

			pre = pre.next
		
		if head2 != None:
			pre.next = head2

		return helper.next