给定一个已排序的双向链表和一个整数X ，任务是打印双向链表中总和为X 的所有四元组。
例子：
Input: LL: -3 ↔ 1 ↔ 2 ↔ 3 ↔ 5 ↔ 6, X = 7
Output:
-3 2 3 5
-3 3 1 6
Explanation: The quadruplets having sum 7( = X) are: {-3, 2, 3, 5}, {-3, 3, 1, 6}.
Input: LL: -2 ↔ -1 ↔ 0 ↔ 0 ↔ 1 ↔ 2, X = 0
Output:
-2 -1 1 2
-2 0 0 2
-1 0 0 1
方法：给定的问题可以通过使用本文中讨论的思想使用 4 指针技术来解决。请按照以下步骤解决问题：
- 初始化四个变量，先说作为双向链表的开始即； first = head ，第二个到第一个指针的下一个，第三个到第二个指针的下一个，第四个到存储排序双向链表中所有 4 个元素的双向链表的最后一个节点。
- 迭代一个循环，直到第一个节点和第四个节点不为NULL ，并且它们不相等并且这两个节点不相互交叉并执行以下步骤：
- 将第二个指针初始化为下一个 第一个。
- 迭代一个循环，直到第二个和第四个节点不为NULL ，它们不相等并且不相互交叉。
- 初始化一个变量，比如sum as (X – (first→data + second→data)) ，将第三个指针指向第二个指针的下一个，并取另一个初始化为最后一个节点的临时指针，即指针第四个。
- 迭代一个循环，而temp和third不为NULL ，它们不相等并且不相互交叉
- 如果总和的值是third→data + temp→data ，则打印四元组并将第三个指针增加到当前第三个指针的下一个，并将temp增加到当前 temp 的前一个。
- 如果 sum的值小于第三个→data + temp→data ，则增加第三个指针，即第三个 = 第三个→next 。
- 否则，递减临时指针，即temp = temp→prev 。
- 将第二个指针移动到下一个指针。
- 将第一个指针移动到下一个指针。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Structure of node of a doubly
// linked list
struct Node {
int data;
struct Node *next, *prev;
};
// Function to insert a new node at
// the beginning of the doubly linked
// list
void insert(struct Node** head, int data)
{
// Allocate the node
struct Node* temp = new Node();
// Fill in the data value
temp->data = data;
temp->next = temp->prev = NULL;
if ((*head) == NULL)
(*head) = temp;
else {
temp->next = *head;
(*head)->prev = temp;
(*head) = temp;
}
}
// Function to print the quadruples
// having sum equal to x
void PrintFourSum(struct Node* head, int x)
{
// First pointer to the head node
struct Node* first = head;
// Pointer to point to the second
// node for the required sum
struct Node* second;
// Pointer to point to the third
// node for the required sum
struct Node* third;
// Fourth points to the last node
struct Node* fourth = head;
// Update the fourth pointer to
// the end of the DLL
while (fourth->next != NULL) {
fourth = fourth->next;
}
// Node to point to the fourth node
// of the required sum
struct Node* temp;
while (first != NULL
&& fourth != NULL
&& first != fourth
&& fourth->next != first) {
// Point the second node to the
// second element of quadruple
second = first->next;
while (second != NULL
&& fourth != NULL
&& second != fourth
&& fourth->next != second) {
int reqsum = x - (first->data
+ second->data);
// Points to the 3rd element
// of quadruple
third = second->next;
// Points to the tail of the DLL
temp = fourth;
while (third != NULL && temp != NULL
&& third != temp
&& temp->next != third) {
// Store the current sum
int twosum = third->data
+ temp->data;
// If the sum is equal,
// then print quadruple
if (twosum == reqsum) {
cout << "(" << first->data
<< ", "
<< second->data
<< ", "
<< third->data
<< ", "
<< temp->data
<< ")\n";
third = third->next;
temp = temp->prev;
}
// If twosum is less than
// the reqsum then move the
// third pointer to the next
else if (twosum < reqsum) {
third = third->next;
}
// Otherwise move the fourth
// pointer to the previous
// of the fourth pointer
else {
temp = temp->prev;
}
}
// Move to the next of
// the second pointer
second = second->next;
}
// Move to the next of
// the first pointer
first = first->next;
}
}
// Driver Code
int main()
{
struct Node* head = NULL;
insert(&head, 2);
insert(&head, 1);
insert(&head, 0);
insert(&head, 0);
insert(&head, -1);
insert(&head, -2);
int X = 0;
PrintFourSum(head, X);
return 0;
} Python3
# Python3 program for the above approach
# Structure of node of a doubly
# linked list
class Node:
def __init__(self, d):
self.data = d
self.left = None
self.right = None
# Function to insert a new node at
# the beginning of the doubly linked
# list
def insert(head, data):
# Allocate the node
temp = Node(data)
# Fill in the data value
temp.data = data
temp.next = temp.prev = None
if (head == None):
head = temp
else:
temp.next = head
head.prev = temp
head = temp
return head
# Function to print the quadruples
# having sum equal to x
def PrintFourSum(head, x):
# First pointer to the head node
first = head
# Pointer to point to the second
# node for the required sum
second = None
# Pointer to point to the third
# node for the required sum
third = None
# Fourth points to the last node
fourth = head
# Update the fourth pointer to
# the end of the DLL
while (fourth.next != None):
fourth = fourth.next
# Node to point to the fourth node
# of the required sum
temp = None
while (first != None and
fourth != None and
first != fourth and
fourth.next != first):
# Point the second node to the
# second element of quadruple
second = first.next
while (second != None and
fourth != None and
second != fourth and
fourth.next != second):
reqsum = x - (first.data +
second.data)
# Points to the 3rd element
# of quadruple
third = second.next
# Points to the tail of the DLL
temp = fourth
while (third != None and temp != None and
third != temp and temp.next != third):
# Store the current sum
twosum = third.data + temp.data
# If the sum is equal,
# then print quadruple
if (twosum == reqsum):
print("(" + str(first.data) +
", " + str(second.data) +
", " + str(third.data) +
", " + str(temp.data) + ")")
third = third.next
temp = temp.prev
# If twosum is less than
# the reqsum then move the
# third pointer to the next
elif (twosum < reqsum):
third = third.next
# Otherwise move the fourth
# pointer to the previous
# of the fourth pointer
else:
temp = temp.prev
# Move to the next of
# the second pointer
second = second.next
# Move to the next of
# the first pointer
first = first.next
# Driver Code
if __name__ == '__main__':
head = None
head = insert(head, 2)
head = insert(head, 1)
head = insert(head, 0)
head = insert(head, 0)
head = insert(head, -1)
head = insert(head, -2)
X = 0
PrintFourSum(head, X)
# This code is contributed by mohit kumar 29输出：
(-2, -1, 1, 2)
(-2, 0, 0, 2)
(-1, 0, 0, 1)时间复杂度： O(N 3 )
辅助空间： O(1)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live