21. Merge Two Sorted Lists

博客围绕合并两个有序链表展开,给出题目链接https://leetcode.com/problems/merge-two-sorted-lists/ ,介绍了代码及思路。思路采用遍历思维,通过比较元素大小,将较小元素放入新链表,指针继续前移。

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题目链接:https://leetcode.com/problems/merge-two-sorted-lists/

代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if not l2 :
            return l1
        
        if not l1 :
            return l2 
        new_node=ListNode(0)
        head = new_node
        while l1 and l2:
            if l1.val < l2.val :
                new_node.next = l1
                l1 = l1.next
            else :
                new_node.next = l2
                l2 = l2.next
            new_node=new_node.next
        if l1 == None and l2 != None :
            new_node.next=l2
        if l1 != None and l2 == None :
            new_node.next=l1
        
        return head.next

思路详解

在这里插入图片描述
遍历的思维,进行比较,若比较小,给新链表,然后指针继续往前。

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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