23. Merge k Sorted Lists-优快云博客

本文链接：https://blog.youkuaiyun.com/csg3140100993/article/details/102789728

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;
    }
};