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。通过初始化指针和迭代比较节点值,确保合并后的链表保持排序状态。

【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.

【思路】

经典链表题,注意p,q的初始化。

【代码】

/**
 * 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 *p,*q;
        ListNode *tmp,*head;
        if(l1==NULL&&l2==NULL) return NULL;
        if(l1==NULL) return l2;
        if(l2==NULL) return l1;
        p=l1;
        q=l2;
        if(p->val>q->val){
            head=q;
            q=q->next;
        }
        else{
            head=p;
            p=p->next;
        }
        tmp=head;
        while(p!=NULL&&q!=NULL){
            if(p->val>q->val){
                tmp->next=q;
                tmp=q;
                q=q->next;
            }
            else{
                tmp->next=p;
                tmp=p;
                p=p->next;
            }
            
        }
        if(p!=NULL){
            while(p!=NULL){
                tmp->next=p;
                tmp=p;
                p=p->next;
            }
        }
        if(q!=NULL){
            while(q!=NULL){
                tmp->next=q;
                tmp=q;
                q=q->next;
            }
        }
        return head;
    }
};


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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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