Add solution for neetcode: merge k sorted linked lists

neetcode/linked_list/merge_k_sorted_ll/README.md

@@ -0,0 +1,25 @@
+Merge K Sorted Linked Lists
+You are given an array of k linked lists lists, where each list is sorted in ascending order.
+
+Return the sorted linked list that is the result of merging all of the individual linked lists.
+
+Example 1:
+
+Input: lists = [[1,2,4],[1,3,5],[3,6]]
+
+Output: [1,1,2,3,3,4,5,6]
+Example 2:
+
+Input: lists = []
+
+Output: []
+Example 3:
+
+Input: lists = [[]]
+
+Output: []
+Constraints:
+
+0 <= lists.length <= 1000
+0 <= lists[i].length <= 100
+-1000 <= lists[i][j] <= 1000

neetcode/linked_list/merge_k_sorted_ll/solution.py

@@ -0,0 +1,55 @@
+from typing import List, Optional
+
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def merge2lists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        dummy = node = ListNode()
+        while l1 and l2:
+            if l1.val < l2.val:
+                node.next = l1
+                l1 = l1.next
+            else:
+                node.next = l2
+                l2 = l2.next
+            node = node.next
+
+        node.next = l1 or l2
+
+        return dummy.next
+
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+        if len(lists) == 0 or not lists:
+            return None
+        if len(lists) == 1:
+            return lists[0]
+        middle = int(len(lists) / 2)
+        breakpoint()
+        left_side = self.mergeKLists(lists[0:middle])
+        right_side = self.mergeKLists(lists[middle:])
+
+        return self.merge2lists(left_side, right_side)
+
+
+if __name__ == "__main__":
+    s = Solution()
+    e3 = ListNode(3, None)
+    e2 = ListNode(2, e3)
+    e1 = ListNode(1, e2)
+
+    k3 = ListNode(5, None)
+    k2 = ListNode(3, k3)
+    k1 = ListNode(2, k2)
+    k0 = ListNode(0, k1)
+
+    s= Solution()
+    k = s.mergeKLists([e1, k0])
+    while k:
+        print(f"{k.val}")
+        k = k.next