leetcode 堆排序_leetcode合并两个有序数组

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给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

示例 1：

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2：

输入：lists = []
输出：[]
示例 3：

输入：lists = [[]]
输出：[]
 

提示：

k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4

题解

1. 暴力，每次找出所有链表中最小的数，然后加入到链表中
   时间复杂度O(mkk) 其中k为一共有多少个序列，m为序列的平均长度

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
class Solution { 
   
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) { 
   
        ListNode *t = new ListNode(0,NULL);
        ListNode *h = t;
        int pox = -1,val;
        int num = 0;
        for(int i = 0;i < lists.size();i ++){ 
   
            if(lists[i] != NULL)num ++;
        }
        while(num != 0){ 
   
            val = 0,pox = -1;
            for(int i = 0;i < lists.size();i ++){ 
   
                if((pox == -1 && lists[i] != NULL)|| (lists[i] != NULL && lists[i]->val < val))pox = i,val = lists[i]->val;
            }
            t->next = new ListNode(val,NULL);
            t = t->next;
            lists[pox] = lists[pox]->next;
            if(lists[pox] == NULL)num -- ;
        }
        return h->next;
    }
};

2. 分治
   时间复杂度O(mklogk)

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
class Solution { 
   
public:
    ListNode * div(int l,int r,vector<ListNode *>&lists){ 
   
        if(l == r){ 
   
            return lists[l];
        }
        int mid = (l + r) >> 1;
        ListNode *lp = div(l,mid,lists),*rp = div(mid + 1,r,lists);
        ListNode *T = new ListNode(0,NULL);
        ListNode *t = T;
        int val;
        while(lp && rp){ 
   
            T->next = new ListNode(0,NULL);
            if(lp->val < rp->val){ 
   
                T->next->val = lp->val;
                lp = lp->next;
            }
            else{ 
   
                T->next->val = rp->val;
                rp = rp->next;
            }
            T = T->next;
        }
        while(lp){ 
   
            T->next = new ListNode(lp->val,NULL);
            lp = lp->next;
            T = T->next;
        }
        while(rp){ 
   
            T->next = new ListNode(rp->val,NULL);
            rp = rp->next;
            T = T->next;
        }
        return t->next;

    }
    ListNode* mergeKLists(vector<ListNode*>& lists) { 
   
        vector<ListNode*>res;
        if(lists.size() == 0)return NULL;
        return div(0,lists.size() - 1,lists);
    }
};

3. 堆
   时间复杂度O(mklogk)

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
class Solution { 
   
public:
    struct node{ 
   
        int pox,val;
        node(int pox,int val):pox(pox),val(val){ 
   }
        bool operator<(const node &a)const{ 
   
            return val > a.val;
        }
    };
    ListNode* mergeKLists(vector<ListNode*>& lists){ 
   
        priority_queue<node,vector<node>> pq;
        int num = 0;
        for(int i = 0;i < lists.size();i ++)
            if(lists[i] != NULL)
                pq.push(node(i,lists[i]->val)),num ++;
        ListNode *T = new ListNode(0,NULL);
        ListNode *h = T;
        while(!pq.empty()){ 
   
            T->next = new ListNode(0,NULL);
            T->next->val = pq.top().val;
            int pox = pq.top().pox;
            pq.pop();
            lists[pox] = lists[pox] -> next;
            if(lists[pox] != NULL){ 
   
                pq.push(node(pox,lists[pox]->val));
            }
            T = T->next;
        }
        return h->next;

    }
};