Merge k Sorted Lists in Python

PythonServer Side ProgrammingProgramming

Suppose we have some lists, these are sorted. We have to merge these lists into one list. To solve this, we will use the heap data structure. So if the lists are [1,4,5], [1,3,4], [2,6], then the final list will be [1,1,2,3,4,4,5,6].

To solve this, we will follow these steps −

  • make one heap

  • for each linked list l in lists −

    • if is in not 0, then insert I into a heap

  • res := null and res_next := null

  • Do one infinite loop −

    • temp := min of heap

    • if heap has no element, then return res

    • if res is 0, then

      • res := temp, res_next := temp

      • temp := next element of temp

      • if temp is not zero, then insert temp into heap

      • next of res := null

    • otherwise −

      • next of res_next := temp, temp := next of temp, res_next := next of res_next

      • if temp is not null, then insert temp into heap

      • next of res_next := null

Example

Let us see the following implementation to get a better understanding −

 Live Demo

class ListNode:
   def __init__(self, data, next = None):
      self.val = data
      self.next = next
def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
         ptr = ptr.next
      ptr.next = ListNode(element)
   return head
def print_list(head):
   ptr = head
   print('[', end = "")
   while ptr:
      print(ptr.val, end = ", ")
      ptr = ptr.next
   print(']')
class Heap:
   def __init__(self):
      self.arr = []
   def print_heap(self):
      res = " "
      for i in self.arr:
         res += str(i.val) + " "
      print(res)
   def getVal(self,i):
      return self.arr[i].val
   def parent(self,i):
      return (i-1)//2
   def left(self,i):
      return (2*i + 1)
   def right(self,i):
      return (2*i + 2)
   def insert(self,value):
      self.arr.append(value)
      n = len(self.arr)-1
      i = n
      while i != 0 and
self.arr[i].val<self.arr[self.parent(i)].val:
         self.arr[i],self.arr[self.parent(i)] = self.arr[self.parent(i)],self.arr[i]
         i = self.parent(i)
   def heapify(self,i):
      left = self.left(i)
      right = self.right(i)
      smallest = i
      n= len(self.arr)
      if left<n and self.getVal(left)<self.getVal(smallest): smallest = left
      if right <n and self.getVal(right)<self.getVal(smallest): smallest = right
      if smallest!=i:
         self.arr[i],self.arr[smallest] = self.arr[smallest],self.arr[i]
         self.heapify(smallest)
   def extractMin(self):
      n = len(self.arr)
      if n==0:
         return '#'
      if n== 1:
         temp =self.arr[0]
         self.arr.pop()
         return temp
      root = self.arr[0]
      self.arr[0] = self.arr[-1]
      self.arr.pop()
      self.heapify(0)
      return root
class Solution(object):
   def mergeKLists(self, lists):
      heap = Heap()
      for i in lists:
         if i:
            heap.insert(i)
      res = None
      res_next = None
      while True:
         temp = heap.extractMin()
         if temp == "#":
            return res
         if not res:
            res = temp
            res_next = temp
            temp = temp.next
            if temp:
               heap.insert(temp)
            res.next = None
      else:
         res_next.next = temp
         temp = temp.next
         res_next=res_next.next
         if temp:
            heap.insert(temp)
         res_next.next = None
ob = Solution()
lists = [[1,4,5],[1,3,4],[2,6]]
lls = []
for ll in lists:
   l = make_list(ll)
   lls.append(l)
print_list(ob.mergeKLists(lls))

Input

[[1,4,5],[1,3,4],[2,6]]

Output

[1, 1, 2, 3, 4, 4, 5, 6, ]
raja
Published on 26-May-2020 14:56:57
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