Merge Two Sorted Lists

本文详细介绍了如何将两个已排序的链表合并为一个排序后的链表,并提供了具体的代码实现。通过比较两个链表中的元素,将较小的元素依次加入到新链表中,直至一个链表为空,然后将另一个链表直接连接到新链表的末尾。

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

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思路:

   新建一个链表,然后比较两个链表中的元素值,把较小的那个链到新链表中,然后较小链表与另一个链表的额值继续进行比较,跟合并数组差不多,由于两个输入链表的长度可能不同,所以最终会有一个链表先完成插入所有元素,则直接另一个未完成的链表直接链入新链表的末尾。


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* newnode=new ListNode(NULL);
        ListNode* head=newnode;//保存头结点
        while(l1!=NULL && l2!=NULL)
        {
            if(l1->val<l2->val)
            {
                newnode->next=l1;
                l1=l1->next;//进行l1链表的 下一个节点与l2比较
            }
            else
            {
                newnode->next=l2;
                l2=l2->next;
            }
            newnode=newnode->next;//保持最前方的节点
        }
        if(l1==NULL) newnode->next=l2;
        if(l2==NULL) newnode->next=l1;
        return head->next; //因为头结点是有元素的,所以链表的头结点是head->next
    }
};


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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