Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Merge k Sorted Lists in Python
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 −
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, ]
677 Views
