合并两个没有重复的排序链表
原文:https://www.geeksforgeeks.org/merge-two-sorted-linked-list-without-duplicates/
合并大小为n1
和n2
的两个排序链表。 在最终排序的链表中,两个链表中的重复项应仅出现一次。
例子:
Input : list1: 1->1->4->5->7
list2: 2->4->7->9
Output : 1 2 4 5 7 9
方法:以下是步骤:
-
以排序方式合并两个排序的链表。 请参阅此帖子的递归方法。 令最终获得的列表为头。
// C++ implementation to merge two sorted linked list
// without duplicates
#include <bits/stdc++.h>
using namespace std;
// structure of a node
struct Node {
int data;
Node* next;
};
// function to get a new node
Node* getNode(int data)
{
// allocate space
Node* temp = (Node*)malloc(sizeof(Node));
// put in data
temp->data = data;
temp->next = NULL;
return temp;
}
// function to merge two sorted linked list
// in a sorted manner
Node* sortedMerge(struct Node* a, struct Node* b)
{
Node* result = NULL;
/* Base cases */
if (a == NULL)
return (b);
else if (b == NULL)
return (a);
/* Pick either a or b, and recur */
if (a->data <= b->data) {
result = a;
result->next = sortedMerge(a->next, b);
}
else {
result = b;
result->next = sortedMerge(a, b->next);
}
return (result);
}
/* The function removes duplicates from a sorted list */
void removeDuplicates(Node* head)
{
/* Pointer to traverse the linked list */
Node* current = head;
/* Pointer to store the next pointer of a node to be deleted*/
Node* next_next;
/* do nothing if the list is empty */
if (current == NULL)
return;
/* Traverse the list till last node */
while (current->next != NULL) {
/* Compare current node with next node */
if (current->data == current->next->data) {
/* The sequence of steps is important*/
next_next = current->next->next;
free(current->next);
current->next = next_next;
}
else /* This is tricky: only advance if no deletion */
{
current = current->next;
}
}
}
// function to merge two sorted linked list
// without duplicates
Node* sortedMergeWithoutDuplicates(Node* head1, Node* head2)
{
// merge two linked list in sorted manner
Node* head = sortedMerge(head1, head2);
// remove duplicates from the list 'head'
removeDuplicates(head);
return head;
}
// function to print the linked list
void printList(Node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}
// Driver program to test above
int main()
{
// head1: 1->1->4->5->7
Node* head1 = getNode(1);
head1->next = getNode(1);
head1->next->next = getNode(4);
head1->next->next->next = getNode(5);
head1->next->next->next->next = getNode(7);
// head2: 2->4->7->9
Node* head2 = getNode(2);
head2->next = getNode(4);
head2->next->next = getNode(7);
head2->next->next->next = getNode(9);
Node* head3;
head3 = sortedMergeWithoutDuplicates(head1, head2);
printList(head3);
return 0;
}
输出:
1 2 4 5 7 9
时间复杂度:O(n1 + n2)
。
辅助空间:O(1)
。
练习:获得最终排序的链表,一次遍历两个列表时没有重复项。
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