23. Merge k Sorted Lists

本文介绍了一种高效算法,用于合并多个已排序的链表成一个单一的排序链表。通过使用优先队列和双指针技术,该算法能够有效地处理大量数据,降低复杂度。文章还提供了详细的代码实现和解析。

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23. Merge k Sorted Lists

Hard

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

Accepted

478,157

Submissions

1,303,920

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int num=lists.size();
        ListNode* temp;
        if(num==0) return NULL;
        else{temp=lists[0];}
        for(int i=1;i<num;i++){
        	temp=mergeTwoLists(temp,lists[i]);
        }
        return temp;
    }
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* head=new ListNode(0);//dummy head
        
        ListNode* p=head;
        while(l1!=NULL&&l2!=NULL){
        	if(l1->val<=l2->val){
        		p->next=l1;
        		l1=l1->next;
        		p=p->next;
                p->next=NULL;
        	}else{
        		p->next=l2;
        		l2=l2->next;
        		p=p->next;
                p->next=NULL;
        	}
        }
        if(l1!=NULL) p->next=l1;
        else if(l2!=NULL) p->next=l2;
        head=head->next;
        return head;
    }
};

这道题用优先队列做比较好,下面为参考代码:

struct cmp {
    bool operator()(ListNode * x, ListNode * y) {//重载运算符
        return x->val > y->val;
    }
};
 
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<ListNode*, vector<ListNode*>, cmp> q;
        for(int i = 0; i < lists.size(); i++)
            if(lists[i] != NULL)
                q.push(lists[i]);
        ListNode *dummy = new ListNode(0);
        ListNode *cur = dummy, *tmp = NULL;
        while(!q.empty()) {//比较头元素
            tmp = q.top();
            q.pop();
            cur->next = tmp;
            cur = cur->next;
            if(tmp->next != NULL)
                q.push(tmp->next);
        }
        cur = dummy->next;
        delete dummy;
        return cur;
    }
};

 

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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