165.Merge Two Sorted Lists-合并两个排序链表(容易题)

本文介绍了一种合并两个已排序链表的方法,通过迭代比较两个链表节点的值,并选择较小值连接到新链表中,直至一个链表遍历完成,然后直接连接剩余链表,最终形成一个新的排序链表。

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合并两个排序链表

  1. 题目

    将两个排序链表合并为一个新的排序链表

  2. 样例

    给出1->3->8->11->15->null,2->null,返回 1->2->3->8->11->15->null。

  3. 题解

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode lastNode = dummy;

        while (l1 != null && l2 != null) 
        {
            if (l1.val < l2.val) 
            {
                lastNode.next = l1;
                l1 = l1.next;
            } 
            else 
            {
                lastNode.next = l2;
                l2 = l2.next;
            }
            lastNode = lastNode.next;
        }
        lastNode.next = l1 == null ? l2 : l1;

        return dummy.next;
    }
}

Last Update 2016.9.7

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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