链表问题-----排序

Posted 2021-01-10 lee-yl

一、题目：将单向链表按某值划分为左边小、中间相等、右边大的形式

简单的思路：时间O（N），空间O（N）

采用一个数组来存储链表中的值，然后对该数组进行快排，然后再连成链表，快排思想如下图所示：

代码：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
#快排思想
def partition(arr,pivot):
    small = -1
    big = len(arr)
    index = 0
    while index != big:
        if arr[index] < pivot:
            small += 1
            arr[small] , arr[index] = arr[index] , arr[small]
            index += 1
        elif arr[index] == pivot:
            index += 1
        else:
            big -= 1
            arr[index] , arr[big] = arr[big] , arr[index]
    return arr
#链表操作
def listSort(head,pivot):
    if not head:
        return head
#将链表存储在数组中
    arr = []
    while head:
        arr.append(head.value)
        head = head.next
    arr = partition(arr,pivot) 
#创建链表   
    node = Node(arr[0])
    p = node
    for i in range(1,len(arr)):
        p.next = Node(arr[i])
        p = p.next
    return node

head = Node(9)
head.next = Node(0)
head.next.next = Node(4)
head.next.next.next = Node(3)
head.next.next.next.next = Node(1)
pivot = 3
node = listSort(head,pivot)

进阶思想：时间O（N），空间O（1）

 

代码：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def listPartition(head,pivot):
    if not head:
        return head
    sH = None
    sT = None
    eH = None
    eT = None
    bH = None
    bT = None   
    while head:
        next = head.next
        head.next = None
        if head.value < pivot:
            if not sH:
                sH = head
                sT = head
            else:
                sT.next = head
                sT = head
        elif head.value == pivot:
            if not eH:
                eH = head
                eT = head
            else:
                eT.next = head
                eT = head
        else:
            if bH == None:
                bH = head
                bT = head
            else:
                bT.next = head
                bT = head
        head = next
    
    if sT:
        sT.next = eH
        eT = eT if eT else sT
    if eT:
        eT.next = bH
    return sH if sH else eH if eH else bH

arr = [9,1,4,5,6,3,8]
head = Node(arr[0])
p = head
for i in range(1,len(arr)):
    p.next = Node(arr[i])
    p = p.next
pivot = 5
res = listPartition(head,pivot)