给定一个单链表，该单链表表示一个数字，其中每个节点仅包含一个数字[0-9] 。任务是在给定的链表所代表的数字上加1并打印新的链表。
例子：
Input: 1 -> 2 -> 9 -> 9 -> NULL
Output:
Original list is : 1 2 9 9
Resultant list is : 1 3 0 0
Input: 9 -> 9 -> 9 -> 9 -> NULL
Output:
Original list is : 9 9 9 9
Resultant list is : 1 0 0 0 0
方法：本文讨论了上述问题的先前实现。但是，一种实现方式要求反向链接列表，而另一种则使用递归。这里讨论了O(1)空间复杂度解决方案，该解决方案不需要反向链接列表。
这个问题的主要焦点在数字9上，它会创建所有更改，否则对于其他数字，我们只需要将其值增加1，但是如果我们将节点的值更改为9，则会产生一个进位，然后必须将其传递通过链接列表。
在链接列表中找到不等于9的最后一个节点。现在有三种情况：
- 如果没有这样的节点，即每个节点的值为9，则新的链表将包含全0，并在链表的开头插入一个1。
- 如果最右边的节点(不等于9)是链表中的最后一个节点，则将该节点加1并返回链表的头。
- 如果该节点不是最后一个节点，即其后的每个节点等于9，则将1添加到当前节点，并将其后的所有节点都更改为0。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Node of the linked list
struct Node {
int data;
Node* next;
};
// Function to create a new node
Node* create_Node(int data)
{
Node* temp = new Node();
temp->data = data;
temp->next = NULL;
return temp;
}
// Function to print the linked list
void print(Node* head)
{
Node* temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
cout << endl;
}
// Function to add one to a number
// represented as linked list
Node* addOne(Node* head)
{
// To store the last node in the linked
// list which is not equal to 9
Node* last = NULL;
Node* cur = head;
// Iterate till the last node
while (cur->next != NULL) {
if (cur->data != 9) {
last = cur;
}
cur = cur->next;
}
// If last node is not equal to 9
// add 1 to it and return the head
if (cur->data != 9) {
cur->data++;
return head;
}
// If list is of the type 9 -> 9 -> 9 ...
if (last == NULL) {
last = new Node();
last->data = 0;
last->next = head;
head = last;
}
// For cases when the righmost node which
// is not equal to 9 is not the last
// node in the linked list
last->data++;
last = last->next;
while (last != NULL) {
last->data = 0;
last = last->next;
}
return head;
}
// Driver code
int main()
{
Node* head = create_Node(1);
head->next = create_Node(2);
head->next->next = create_Node(9);
head->next->next->next = create_Node(9);
cout << "Original list is : ";
print(head);
head = addOne(head);
cout << "Resultant list is : ";
print(head);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Node of the linked list
static class Node
{
int data;
Node next;
};
// Function to create a new node
static Node create_Node(int data)
{
Node temp = new Node();
temp.data = data;
temp.next = null;
return temp;
}
// Function to print the linked list
static void print(Node head)
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
// Function to add one to a number
// represented as linked list
static Node addOne(Node head)
{
// To store the last node in the linked
// list which is not equal to 9
Node last = null;
Node cur = head;
// Iterate till the last node
while (cur.next != null)
{
if (cur.data != 9)
{
last = cur;
}
cur = cur.next;
}
// If last node is not equal to 9
// add 1 to it and return the head
if (cur.data != 9)
{
cur.data++;
return head;
}
// If list is of the type 9 . 9 . 9 ...
if (last == null)
{
last = new Node();
last.data = 0;
last.next = head;
head = last;
}
// For cases when the righmost node which
// is not equal to 9 is not the last
// node in the linked list
last.data++;
last = last.next;
while (last != null)
{
last.data = 0;
last = last.next;
}
return head;
}
// Driver code
public static void main(String[] args)
{
Node head = create_Node(1);
head.next = create_Node(2);
head.next.next = create_Node(9);
head.next.next.next = create_Node(9);
System.out.print("Original list is : ");
print(head);
head = addOne(head);
System.out.print("Resultant list is : ");
print(head);
}
}
// This code is contributed
// by PrinciRaj1992Python3
# A Python3 implementation of the approach
# Node of the linked list
class Node():
def __init__(self):
self.data = None
self.next = None
# Function to create a new node
def create_Node(data):
temp = Node()
temp.data = data
temp.next = None
return temp
# Function to print the linked list
def printList(head):
temp = head
while (temp != None):
print(temp.data, end = ' ')
temp = temp.next
print()
# Function to add one to a number
# represented as linked list
def addOne(head):
# To store the last node in the
# linked list which is not equal to 9
last = None
cur = head
# Iterate till the last node
while(cur.next != None):
if(cur.data != 9):
last = cur
cur = cur.next
# If last node is not equal to 9
# add 1 to it and return the head
if(cur.data != 9):
cur.data += 1
return head
# If list is of the type 9 -> 9 -> 9 ...
if(last == None):
last = Node()
last.data = 0
last.next = head
head = last
# For cases when the rightmost node which
# is not equal to 9 is not the last
# node in the linked list
last.data += 1
last = last.next
while(last != None):
last.data = 0
last = last.next
return head
# Driver code
if __name__=='__main__':
head = create_Node(1)
head.next = create_Node(2)
head.next.next = create_Node(9)
head.next.next.next = create_Node(9)
print("Original list is : ", end = "")
printList(head)
head = addOne(head)
print("Resultant list is : ", end = "")
printList(head)
# This code is contributed by Yashyasvi AgarwalC#
// C# implementation of the approach
using System;
class GFG
{
// Node of the linked list
public class Node
{
public int data;
public Node next;
};
// Function to create a new node
static Node create_Node(int data)
{
Node temp = new Node();
temp.data = data;
temp.next = null;
return temp;
}
// Function to print the linked list
static void print(Node head)
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
// Function to add one to a number
// represented as linked list
static Node addOne(Node head)
{
// To store the last node in the linked
// list which is not equal to 9
Node last = null;
Node cur = head;
// Iterate till the last node
while (cur.next != null)
{
if (cur.data != 9)
{
last = cur;
}
cur = cur.next;
}
// If last node is not equal to 9
// add 1 to it and return the head
if (cur.data != 9)
{
cur.data++;
return head;
}
// If list is of the type 9 . 9 . 9 ...
if (last == null)
{
last = new Node();
last.data = 0;
last.next = head;
head = last;
}
// For cases when the righmost node which
// is not equal to 9 is not the last
// node in the linked list
last.data++;
last = last.next;
while (last != null)
{
last.data = 0;
last = last.next;
}
return head;
}
// Driver code
public static void Main(String[] args)
{
Node head = create_Node(1);
head.next = create_Node(2);
head.next.next = create_Node(9);
head.next.next.next = create_Node(9);
Console.Write("Original list is : ");
print(head);
head = addOne(head);
Console.Write("Resultant list is : ");
print(head);
}
}
// This code is contributed by 29AjayKumar输出：
Original list is : 1 2 9 9
Resultant list is : 1 3 0 0时间复杂度： O(N)
空间复杂度： O(1)