LeetCode-23. Merge k Sorted Lists [C++][Java]

LeetCode-23. Merge k Sorted Listshttps://leetcode.com/problems/merge-k-sorted-lists/submissions/

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 <= 104
• 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 will not exceed 10^4.

【C++】

构造最小堆

/**
 * 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& lists) {
        if (lists.empty()) return nullptr;
        auto cmp = [](const ListNode* l1, const ListNode* l2) {
            return l1->val > l2->val;
        };
        priority_queue, decltype(cmp)> q(cmp);
        for (ListNode* list: lists) {
            if (list) {
                q.push(list);
            }
        }
        ListNode* dummy = new ListNode(0), *cur = dummy;
        while (!q.empty()) {
            cur->next = q.top();
            q.pop();
            cur = cur->next;
            if (cur->next) {
                q.push(cur->next);
            }
        }
        return dummy->next;
    }
};

【Java】

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        PriorityQueue q = new PriorityQueue<>((a, b) -> a.val - b.val);
        for (ListNode list: lists) {
            if (list != null) {
                q.offer(list);
            }
        }
        ListNode dummy = new ListNode(0), cur = dummy;
        while (!q.isEmpty()) {
            cur.next = q.poll();
            cur = cur.next;
            if (cur.next != null) {
                q.offer(cur.next);
            }
        }
        return dummy.next;
    }
}

参考文献

【1】STL 最大堆、最小堆的应用_沐沐余风的博客-CSDN博客_c++最小堆stl