LeetCode 23 Merge k Sorted Lists (C++, Python)

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


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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
struct NodeComp {
    bool operator ()(const pair<int, ListNode*> n1, const pair<int, ListNode*> n2) {
        return n1.second->val > n2.second->val;
    }
};
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        ListNode* newHead = NULL;
        ListNode* newTail = NULL;
        if (lists.size() == 0) return newHead;
        priority_queue<pair<int, ListNode*>, vector<pair<int, ListNode*>>, NodeComp> myqueue;
        for (int i = 0; i < lists.size(); ++i) {
            ListNode* head = lists[i];
            if (head) {
                lists[i] = head->next;
                head->next = NULL;
                myqueue.push({i, head});
            }
        }
        while (!myqueue.empty()) {
            pair<int, ListNode*> node = myqueue.top();
            myqueue.pop();
            if (newHead) {
                newTail->next = node.second;
                newTail = node.second;
            } else {
                newHead = newTail = node.second;
            }
            ListNode* head = lists[node.first];
            if (head) {
                lists[node.first] = head->next;
                head->next = NULL;
                myqueue.push({node.first, head});
            }
        }
        return newHead;
    }
};
Solution Python


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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

import heapq
class Solution:
    # @param a list of ListNode
    # @return a ListNode
    def mergeKLists(self, lists):
        newHead = None
        newTail = None
        if not lists: return newHead
        h = []
        for i in range(len(lists)):
            head = lists[i]
            if not head: continue
            lists[i] = head.next
            head.next = None
            heapq.heappush(h, (head.val, i, head))
        while h:
            v, i, n = heapq.heappop(h)
            if newHead:
                newTail.next = n
                newTail = n
            else:
                newHead = newTail = n
            head = lists[i]
            if not head: continue
            lists[i] = head.next
            head.next = None
            heapq.heappush(h, (head.val, i, head))
        return newHead

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