给定一个链接列表0、1和2，对其进行排序。
例子：
Input: 1 -> 1 -> 2 -> 0 -> 2 -> 0 -> 1 -> NULL
Output: 0 -> 0 -> 1 -> 1 -> 1 -> 2 -> 2 -> NULL
Input: 1 -> 1 -> 2 -> 1 -> 0 -> NULL
Output: 0 -> 1 -> 1 -> 1 -> 2 -> NULL
资料来源：Microsoft专访|套装1
可以使用以下步骤对给定的链表进行排序。
- 遍历列表并计算0、1和2的数量。设计数分别为n1，n2和n3。
- 再次遍历该列表，将前n1个节点填充为0，然后将n2个节点填充为1，最后将n3个节点填充为2。
下图是上述方法的模拟：
下面是上述方法的实现：
C++
// C++ Program to sort a linked list 0s, 1s or 2s
#include
using namespace std;
/* Link list node */
class Node
{
public:
int data;
Node* next;
};
// Function to sort a linked list of 0s, 1s and 2s
void sortList(Node *head)
{
int count[3] = {0, 0, 0}; // Initialize count of '0', '1' and '2' as 0
Node *ptr = head;
/* count total number of '0', '1' and '2'
* count[0] will store total number of '0's
* count[1] will store total number of '1's
* count[2] will store total number of '2's */
while (ptr != NULL)
{
count[ptr->data] += 1;
ptr = ptr->next;
}
int i = 0;
ptr = head;
/* Let say count[0] = n1, count[1] = n2 and count[2] = n3
* now start traversing list from head node,
* 1) fill the list with 0, till n1 > 0
* 2) fill the list with 1, till n2 > 0
* 3) fill the list with 2, till n3 > 0 */
while (ptr != NULL)
{
if (count[i] == 0)
++i;
else
{
ptr->data = i;
--count[i];
ptr = ptr->next;
}
}
}
/* Function to push a node */
void push (Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = new Node();
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(Node *node)
{
while (node != NULL)
{
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
/* Driver code*/
int main(void)
{
Node *head = NULL;
push(&head, 0);
push(&head, 1);
push(&head, 0);
push(&head, 2);
push(&head, 1);
push(&head, 1);
push(&head, 2);
push(&head, 1);
push(&head, 2);
cout << "Linked List Before Sorting\n";
printList(head);
sortList(head);
cout << "Linked List After Sorting\n";
printList(head);
return 0;
}
// This code is contributed by rathbhupendra C
// C Program to sort a linked list 0s, 1s or 2s
#include
#include
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
// Function to sort a linked list of 0s, 1s and 2s
void sortList(struct Node *head)
{
int count[3] = {0, 0, 0}; // Initialize count of '0', '1' and '2' as 0
struct Node *ptr = head;
/* count total number of '0', '1' and '2'
* count[0] will store total number of '0's
* count[1] will store total number of '1's
* count[2] will store total number of '2's */
while (ptr != NULL)
{
count[ptr->data] += 1;
ptr = ptr->next;
}
int i = 0;
ptr = head;
/* Let say count[0] = n1, count[1] = n2 and count[2] = n3
* now start traversing list from head node,
* 1) fill the list with 0, till n1 > 0
* 2) fill the list with 1, till n2 > 0
* 3) fill the list with 2, till n3 > 0 */
while (ptr != NULL)
{
if (count[i] == 0)
++i;
else
{
ptr->data = i;
--count[i];
ptr = ptr->next;
}
}
}
/* Function to push a node */
void push (struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct Node *node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
printf("n");
}
/* Driver program to test above function*/
int main(void)
{
struct Node *head = NULL;
push(&head, 0);
push(&head, 1);
push(&head, 0);
push(&head, 2);
push(&head, 1);
push(&head, 1);
push(&head, 2);
push(&head, 1);
push(&head, 2);
printf("Linked List Before Sorting\n");
printList(head);
sortList(head);
printf("Linked List After Sorting\n");
printList(head);
return 0;
} Java
// Java program to sort a linked list of 0, 1 and 2
class LinkedList
{
Node head; // head of list
/* Linked list Node*/
class Node
{
int data;
Node next;
Node(int d) {data = d; next = null; }
}
void sortList()
{
// initialise count of 0 1 and 2 as 0
int count[] = {0, 0, 0};
Node ptr = head;
/* count total number of '0', '1' and '2'
* count[0] will store total number of '0's
* count[1] will store total number of '1's
* count[2] will store total number of '2's */
while (ptr != null)
{
count[ptr.data]++;
ptr = ptr.next;
}
int i = 0;
ptr = head;
/* Let say count[0] = n1, count[1] = n2 and count[2] = n3
* now start traversing list from head node,
* 1) fill the list with 0, till n1 > 0
* 2) fill the list with 1, till n2 > 0
* 3) fill the list with 2, till n3 > 0 */
while (ptr != null)
{
if (count[i] == 0)
i++;
else
{
ptr.data= i;
--count[i];
ptr = ptr.next;
}
}
}
/* Utility functions */
/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Function to print linked list */
void printList()
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data+" ");
temp = temp.next;
}
System.out.println();
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist = new LinkedList();
/* Constructed Linked List is 1->2->3->4->5->6->7->
8->8->9->null */
llist.push(0);
llist.push(1);
llist.push(0);
llist.push(2);
llist.push(1);
llist.push(1);
llist.push(2);
llist.push(1);
llist.push(2);
System.out.println("Linked List before sorting");
llist.printList();
llist.sortList();
System.out.println("Linked List after sorting");
llist.printList();
}
}
/* This code is contributed by Rajat Mishra */Python
# Python program to sort a linked list of 0, 1 and 2
class LinkedList(object):
def __init__(self):
# head of list
self.head = None
# Linked list Node
class Node(object):
def __init__(self, d):
self.data = d
self.next = None
def sortList(self):
# initialise count of 0 1 and 2 as 0
count = [0, 0, 0]
ptr = self.head
# count total number of '0', '1' and '2'
# * count[0] will store total number of '0's
# * count[1] will store total number of '1's
# * count[2] will store total number of '2's
while ptr != None:
count[ptr.data]+=1
ptr = ptr.next
i = 0
ptr = self.head
# Let say count[0] = n1, count[1] = n2 and count[2] = n3
# * now start traversing list from head node,
# * 1) fill the list with 0, till n1 > 0
# * 2) fill the list with 1, till n2 > 0
# * 3) fill the list with 2, till n3 > 0
while ptr != None:
if count[i] == 0:
i+=1
else:
ptr.data = i
count[i]-=1
ptr = ptr.next
# Utility functions
# Inserts a new Node at front of the list.
def push(self, new_data):
# 1 & 2: Allocate the Node &
# Put in the data
new_node = self.Node(new_data)
# 3. Make next of new Node as head
new_node.next = self.head
# 4. Move the head to point to new Node
self.head = new_node
# Function to print linked list
def printList(self):
temp = self.head
while temp != None:
print str(temp.data),
temp = temp.next
print ''
# Driver program to test above functions
llist = LinkedList()
llist.push(0)
llist.push(1)
llist.push(0)
llist.push(2)
llist.push(1)
llist.push(1)
llist.push(2)
llist.push(1)
llist.push(2)
print "Linked List before sorting"
llist.printList()
llist.sortList()
print "Linked List after sorting"
llist.printList()
# This code is contributed by BHAVYA JAINC#
// C# program to sort a linked
// list of 0, 1 and 2
using System;
public class LinkedList
{
Node head; // head of list
/* Linked list Node*/
class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d; next = null;
}
}
void sortList()
{
// initialise count of 0 1 and 2 as 0
int []count = {0, 0, 0};
Node ptr = head;
/* count total number of '0', '1' and '2'
* count[0] will store total number of '0's
* count[1] will store total number of '1's
* count[2] will store total number of '2's */
while (ptr != null)
{
count[ptr.data]++;
ptr = ptr.next;
}
int i = 0;
ptr = head;
/* Let say count[0] = n1, count[1] = n2 and count[2] = n3
* now start traversing list from head node,
* 1) fill the list with 0, till n1 > 0
* 2) fill the list with 1, till n2 > 0
* 3) fill the list with 2, till n3 > 0 */
while (ptr != null)
{
if (count[i] == 0)
i++;
else
{
ptr.data= i;
--count[i];
ptr = ptr.next;
}
}
}
/* Utility functions */
/* Inserts a new Node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Function to print linked list */
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data+" ");
temp = temp.next;
}
Console.WriteLine();
}
/* Driver code */
public static void Main(String []args)
{
LinkedList llist = new LinkedList();
/* Constructed Linked List is 1->2->3->4->
5->6->7->8->8->9->null */
llist.push(0);
llist.push(1);
llist.push(0);
llist.push(2);
llist.push(1);
llist.push(1);
llist.push(2);
llist.push(1);
llist.push(2);
Console.WriteLine("Linked List before sorting");
llist.printList();
llist.sortList();
Console.WriteLine("Linked List after sorting");
llist.printList();
}
}
/* This code is contributed by 29AjayKumar */输出：
Linked List Before Sorting
2 1 2 1 1 2 0 1 0
Linked List After Sorting
0 0 1 1 1 1 2 2 2时间复杂度： O(n)，其中n是链表中的节点数。
辅助空间： O(1)
通过更改链接对0、1、2和2的链接列表进行排序