Last updated on October 10th, 2024 at 01:58 am

Here, We see Merge k Sorted Lists LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

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Merge k Sorted Lists LeetCode Solution

Problem Statement

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: []

1. Merge k Sorted Lists Leetcode Solution C++

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<ListNode*, vector<ListNode*>, compare> q;
        for (auto l : lists) {
            if (l) {
                q.push(l);
            }
        }
        
        ListNode pre(0);
        ListNode *node = &pre;
        while (q.size()) {
            ListNode *top = q.top();
            q.pop();
            node->next = top;
            node = node->next;
            if (top->next) {
                q.push(top->next);
            }
        }
        return pre.next;
    }
    
    struct compare {
        bool operator()(const ListNode* l1, const ListNode* l2) {
            return l1->val > l2->val;
        }
    };
};

2. Merge k Sorted Lists Leetcode Solution Java

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return partion(lists,0,lists.length-1);
}

public static ListNode partion(ListNode[] lists,int s,int e){
    if(s==e)  return lists[s];
    if(s<e){
        int q=(s+e)/2;
        ListNode l1=partion(lists,s,q);
        ListNode l2=partion(lists,q+1,e);
        return merge(l1,l2);
    }else
        return null;
}
public static ListNode merge(ListNode l1,ListNode l2){
    if(l1==null) return l2;
    if(l2==null) return l1;
    if(l1.val<l2.val){
        l1.next=merge(l1.next,l2);
        return l1;
    }else{
        l2.next=merge(l1,l2.next);
        return l2;
    }
}
}

3. Merge k Sorted Lists Leetcode Solution JavaScript

function mergeLists(a, b) {
    const dummy = new ListNode(0);
    let temp = dummy;
     while (a !== null && b !== null) {
         if (a.val < b.val) {
             temp.next = a;
             a = a.next;
         } else {
             temp.next = b;
             b = b.next;
         }
         temp = temp.next;
     }
    if (a !== null) {
        temp.next = a;
    }
    if (b !== null) {
        temp.next = b;
    }
    return dummy.next;
}

var mergeKLists = function(lists) {
    if (lists.length === 0 ) {
        return null;
    }
    while (lists.length > 1) {
        let a = lists.shift();
        let b = lists.shift();
        const h = mergeLists(a, b);
        lists.push(h);
    }
    return lists[0];
};

4. Merge k Sorted Lists Leetcode Solution Python

class Solution(object):
    def mergeKLists(self, lists):
        if not lists:
            return None
        if len(lists) == 1:
            return lists[0]
        mid = len(lists) // 2
        l, r = self.mergeKLists(lists[:mid]), self.mergeKLists(lists[mid:])
        return self.merge(l, r)
    
    def merge(self, l, r):
        dummy = p = ListNode()
        while l and r:
            if l.val < r.val:
                p.next = l
                l = l.next
            else:
                p.next = r
                r = r.next
            p = p.next
        p.next = l or r
        return dummy.next
    
    def merge1(self, l, r):
        if not l or not r:
            return l or r
        if l.val< r.val:
            l.next = self.merge(l.next, r)
            return l
        r.next = self.merge(l, r.next)
        return r