java中array list和linked list区别

Array List 和 Linked List 都是 Java 中常用的集合类，它们的主要区别在于底层数据结构不同。Array List 底层是数组，支持随机访问，但插入和删除元素时需要移动其他元素，效率较低；而 Linked List 底层是链表，插入和删除元素时只需要改变指针指向，效率较高，但随机访问需要遍历链表，效率较低。因此，如果需要频繁进行插入和删除操作，建议使用 Linked List；如果需要频繁进行随机访问操作，建议使用 Array List。

相关问题

Array list和linked list的区别

Array list 是基于数组实现的列表，它具有随机访问和快速插入、删除元素的能力，而 linked list 则是基于链表实现的列表，它具有快速插入、删除元素的能力，但随机访问元素则需要遍历整个链表。因此，在需要频繁插入、删除元素但不需要随机访问的情况下，linked list 更适合，而在需要随机访问元素的情况下，则应该选择 array list。

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list and return it

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).