LeetCode #148 Sort List

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Problem

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

Examples

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put: 4->2->1->3
Output: 1->2->3->4
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Input: -1->5->3->4->0
Output: -1->0->3->4->5

Solution1

Method: Divide and Conquer

Time Complexity: O(n log n)

Space Complexity: O(1)

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        
        first = slow = head
        fast = head
        
        while fast.next and fast.next.next:
            fast = fast.next.next
            slow = slow.next
        
        second = slow.next
        slow.next = None        
        left = self.sortList(first)
        right = self.sortList(second)
        
        return self.merge(left, right)


    def merge(self, left, right):
        if not left or not right:
            return left or right
        
        if left.val > right.val:
            left, right = right, left
        
        head = prev = left
        left = left.next

        while left and right:
            if left.val < right.val:
                left = left.next
            else:
                new_left = prev.next
                prev.next = right
                next_right = right.next
                right.next = new_left
                right = next_right

            prev = prev.next
        
        prev.next = left or right

        return head

Solution2

Method: Divide and Conquer without recursive

Time Complexity: O(n log n)

Space Complexity: O(1)

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class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None: return None
        
        def getSize(head):
            counter = 0
            while(head is not None):
                counter +=1
                head = head.next
            return counter
        
        def split(head, step):
            i = 1
            while (i < step and head):
                head = head.next
                i += 1
                
            if head is None: return None
            #disconnect
            temp, head.next = head.next, None
            return temp
        
        def merge(l, r, head):
            cur = head
            while(l and r):
                if l.val < r.val:
                    cur.next, l = l, l.next
                else:
                    cur.next, r = r, r.next
                cur = cur.next
            
            cur.next = l if l is not None else r
            while cur.next is not None: cur = cur.next
            return cur

        size = getSize(head)
        bs = 1
        dummy = ListNode(0)
        dummy.next = head
        l, r, tail = None, None, None
        while (bs < size):
            cur = dummy.next
            tail = dummy
            while cur:
                l = cur
                r = split(l, bs)
                cur = split(r, bs)
                tail = merge(l, r, tail)
            bs <<= 1
        return dummy.next