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LeetCode 23. Merge k Sorted Lists

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题目

原题在此

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
1->4->5,
1->3->4,
2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

Constraints:

  • k == lists.length
  • 0 <= k <= 10^4
  • 0 <= lists[i].length <= 500
  • -10^4 <= lists[i][j] <= 10^4
  • lists[i] is sorted in ascending order.
  • The sum of lists[i].length won’t exceed $10^4$.

解析

做过TopK这道题的应该对最大堆,优先队列等数据结构有印象. 想看实现细节的可以自己百度.
将lists中的每个节点存入优先队列中, val最小的node出队, 出队node的next入队即可.

代码

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/**
 * 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 Compare {
public:
    bool operator() (ListNode* foo, ListNode* bar) {
        return foo->val > bar->val;
    }
};
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<ListNode*, vector<ListNode*>, Compare> pq;
        for(ListNode* node : lists)
            if(node) pq.push(node);
        ListNode* res = new ListNode();
        ListNode* foo = res;
        while(!pq.empty()) {
            foo->next = pq.top();
            pq.pop();
            foo = foo->next;
            if(foo->next) pq.push(foo->next);
        }
        return res->next;
    }
};