merge two sorted linked list without duplicates

本文介绍了一种算法,在不使用额外空间的情况下,将两个已排序的链表合并,并移除重复元素。

Question:

Given two sorted linked list, and merge them without using extra space (using constant space is allowed). If there exist duplicated items, remove them and leave only one copy.

public static Node mergeSortedListWithouDuplicates(Node h1, Node h2) {
	//one or both of the list(s) is (are) null
	if(h1 == null) return h2;
	if(h2 == null ) return h1;
	Node newHead = (h1.data <= h2.data) ? h1 : h2;
	
	Node small = null;
	Node prev = newHead;
	while (h1 != null && h2 != null) {
		if (h1.data <= h2.data) {
			small = h1;
			h1 = h1.next;
		} else {
			small = h2;
			h2 = h2.next;
		}
		if (prev.data != small.data) {
			prev.next = small;
			prev = small;
		}
	}
	
	//deal with the remaining part of h1 or h2
	while (h2 != null) {
		if (prev.data != h2.data) {
			prev.next = h2;
			prev = h2;
		}
		h2 = h2.next;
	}
	while (h1 != null) {
		if (prev.data != h1.data) {
			prev.next = h1;
			prev = h1;
		}
		h1 = h1.next;
	}
	// important
	prev.next = null;
	return newHead;		
}
blog.youkuaiyun.com/beiyetengqing

To merge k sorted linked lists, one approach is to repeatedly merge two of the linked lists until all k lists have been merged into one. We can use a priority queue to keep track of the minimum element across all k linked lists at any given time. Here's the code to implement this idea: ``` struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; // Custom comparator for the priority queue struct CompareNode { bool operator()(const ListNode* node1, const ListNode* node2) const { return node1->val > node2->val; } }; ListNode* mergeKLists(vector<ListNode*>& lists) { priority_queue<ListNode*, vector<ListNode*>, CompareNode> pq; for (ListNode* list : lists) { if (list) { pq.push(list); } } ListNode* dummy = new ListNode(-1); ListNode* curr = dummy; while (!pq.empty()) { ListNode* node = pq.top(); pq.pop(); curr->next = node; curr = curr->next; if (node->next) { pq.push(node->next); } } return dummy->next; } ``` We start by initializing a priority queue with all the head nodes of the k linked lists. We use a custom comparator that compares the values of two nodes and returns true if the first node's value is less than the second node's value. We then create a dummy node to serve as the head of the merged linked list, and a current node to keep track of the last node in the merged linked list. We repeatedly pop the minimum node from the priority queue and append it to the merged linked list. If the popped node has a next node, we push it onto the priority queue. Once the priority queue is empty, we return the head of the merged linked list. Note that this implementation has a time complexity of O(n log k), where n is the total number of nodes across all k linked lists, and a space complexity of O(k).
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