#21. Merge Two Sorted Lists

本文介绍了一种方法来合并两个已排序的链表,并通过重新连接节点的方式创建一个新的排序链表。该方法适用于计算机科学与数据结构的学习者。

题目描述:

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.

/**
 * 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 *temp=l2,*last_temp=NULL,*random;
        while(l1!=NULL){
            temp = l2;
            while(temp!=NULL&&l1->val>temp->val){
                last_temp = temp;
                temp = temp->next;
            }
            /*if(last_temp!=NULL)
                cout<<"last_temp = "<<last_temp->val<<endl;
            if(temp!=NULL)
                cout<<"temp ="<<temp->val<<endl;
                */
            random = l1;
            l1 = l1->next;
            //插入头 
            if(last_temp==NULL){    
                random->next = l2;
                l2 = random;
            }else if(temp==NULL){//插入尾部 
                last_temp->next = random;
                random->next = NULL;
            }else if(temp!=NULL){//插入中间 
                last_temp->next=random;
                random->next = temp;
            }

            random = l2;
            while(random!=NULL){
                //cout<<random->val<<"  ";
                random = random->next;
            } 
            //cout<<endl;
        }
        return l2;
    }
};
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&#39;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&#39;s value is less than the second node&#39;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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