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Commit 4c1b370

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day 2: more python solutions
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‎README.md‎

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| Current Status| Stats |
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| :------------: | :----------: |
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| Total Problems | 188 |
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| Current Daily Streak| 1 |
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| Last updated | 06/20/2019|
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| Current Daily Streak| 2 |
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| Last updated | 06/21/2019|
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</center>
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@@ -17,12 +17,12 @@
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### LinkedList Problems
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| Problem | Solution |
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| :------------ | :----------: |
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| Find the nth node of linked list from last. |[nthToLastNode.cpp](linked_list_problems/nthToLastNode.cpp) |
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| Add numbers where each digit of the number is represented by node of a linkedlist. Give output as a linked list. | [add_two_numbers_lists.cpp](linked_list_problems/add_two_numbers_lists.cpp)|
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| Find the nth node of linked list from last. |[nthToLastNode.cpp](linked_list_problems/nthToLastNode.cpp), [nth_to_last_node.py](linked_list_problems/nth_to_last_node.py) |
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| Add numbers where each digit of the number is represented by node of a linkedlist. Give output as a linked list. | [add_two_numbers_lists.cpp](linked_list_problems/add_two_numbers_lists.cpp), [add_two_numbers_list.py](linked_list_problems/add_two_numbers_lists.py)|
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| Swap nodes of a linkedlist without swapping data. |[swapNodesWithoutSwappingData.cpp](linked_list_problems/swapNodesWithoutSwappingData.cpp)|
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| Reverse a linked list, iteratively and recursively | [reverseLinkedListIterAndRecurse.cpp](linked_list_problems/reverseLinkedListIterAndRecurse.cpp)|
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| Given a linked list, reverse alternate nodes and append at the end. | [reverseAlternateNodes.cpp](linked_list_problems/reverseAlternateNodes.cpp) |
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| Only given a node pointer, delete the node from the linked list. | [deleteNode.cpp](linked_list_problems/deleteNode.cpp)|
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| Only given a node pointer, delete the node from the linked list. | [deleteNode.cpp](linked_list_problems/deleteNode.cpp)|
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| Delete the entire linkedlist. | [deleteLinkedlist.cpp](linked_list_problems/deleteLinkedlist.cpp)|
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| Print middle node of linkedlist without iterating twice. | [printMiddleNode.cpp](linked_list_problems/printMiddleNode.cpp) | | Detecting and removing a cycle in linkedlist.| [floyedCycleDetection.cpp](linked_list_problems/floyedCycleDetection.cpp)|
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| Determine if a linked list is a pallindrome. | [listPallindrome.cpp](linked_list_problems/listPallindrome.cpp) |

‎linked_list_problems/add_two_numbers_lists.cpp‎

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Original file line numberDiff line numberDiff line change
@@ -103,10 +103,5 @@ int main()
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iterate(list2);
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ListNode *list3 = addLists(list1, list2);
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iterate(list3);
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return 0;
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}
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"""
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We have two numbers represented by a linked list,
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where each node contains a single digit. The digits are stored in reverse order,
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such that the 1's digit is at the head of the list.
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Write a function that adds the two numbers and returns the sum as a linked list.
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Example:
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Given 7->1->6 + 5->9->2. That is, 617 + 295.
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Return 2->1->9. That is 912.
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Given 3->1->5 and 5->9->2, return 8->0->8.
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"""
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class Node:
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"""Representation of a linked list node"""
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def __init__(self, data):
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self.data = data
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self.next = None
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def add_two_numbers_lists(list1, list2):
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""" Adds two numbers represented as a linked list and
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returns the sum as a linked list
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Example:
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7->1->6 + 5->9->2 = 2->1->9
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(617 + 295 = 912)
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Params:
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list1 : list representing first number
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list2 : list representing second number
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"""
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list3 = None
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prev = None
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carry = 0
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while list1 is not None or list2 is not None:
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res = carry
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res += 0 if list1 is None else list1.data
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res += 0 if list2 is None else list2.data
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carry = 1 if res >= 10 else 0
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res = res % 10
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newNode = Node(res)
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if list3 is None:
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list3 = newNode
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else:
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prev.next = newNode
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prev = newNode
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if list1 is not None:
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list1 = list1.next
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if list2 is not None:
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list2 = list2.next
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return list3
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def print_list(head):
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while head:
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if head.next is None:
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print(head.data)
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else:
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print(str(head.data)+ '-->', end = ' ')
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head = head.next
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if __name__ == '__main__':
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print("List 1:")
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list1 = Node(7)
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list1.next = Node(1)
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list1.next.next = Node(6)
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print_list(list1)
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print("List 2:")
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list2 = Node(5)
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list2.next = Node(9)
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list2.next.next = Node(2)
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print_list(list2)
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print("Result List:")
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list3 = add_two_numbers_lists(list1, list2)
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print_list(list3)
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print("Example 2: ")
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print("List 1:")
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list1 = Node(3)
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list1.next = Node(1)
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list1.next.next = Node(5)
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print_list(list1)
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print("List 2:")
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list2 = Node(5)
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list2.next = Node(9)
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list2.next.next = Node(2)
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print_list(list2)
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print("Result List:")
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list3 = add_two_numbers_lists(list1, list2)
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print_list(list3)

‎linked_list_problems/delete_node.py‎

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# Problem: Only given a node of the linked list, delete the node.
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class Node:
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"""Representation of a node of a linked list"""
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def __init__(self, data):
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self.data = data
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self.next = None
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class LinkedList:
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"""Representation of a linked list"""
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def __init__(self):
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self.head = None
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def push(self, data):
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"""Pushes a new node to the linkedlist, the new node becomes head
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Params:
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data of the new node.
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"""
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new_node = Node(data)
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new_node.next = self.head
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self.head = new_node
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def print_list(self):
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"""Prints the linked list"""
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itr = self.head
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while itr:
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if itr.next is None:
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print(itr.data)
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else:
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print(str(itr.data)+ "-->", end=" ")
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itr = itr.next
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def delete_node(self, node):
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"""Deletes a given node from the linked list
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Params:
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node: node to be deleted
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"""
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itr = self.head
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while itr.next != node:
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itr = itr.next
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if itr is not None:
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itr.next = node.next
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node.next = None
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def get_head(self):
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return self.head
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if __name__ == '__main__':
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linkedList = LinkedList()
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linkedList.push(1)
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linkedList.push(2)
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linkedList.push(3)
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linkedList.push(4)
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linkedList.push(4)
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print("List before deleting:")
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linkedList.print_list()
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# Delete second node from the list.
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linkedList.delete_node(linkedList.get_head().next)
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print("List after deleting:")
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linkedList.print_list()
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"""
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Find nth to last node in a linked list.
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"""
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class Node:
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def __init__(self, data):
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"""Representation of a linked list node"""
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self.data = data
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self.next = None
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def nth_to_last(head, n):
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""" Find nth to last node of a linked list
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Params:
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head : head of the linked list
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n: the distance of target node from the end of linkedlist.
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"""
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itr1 = head
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itr2 = head
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i = 1
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# First move the first iterator to nth node from beginning.
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while itr1 and i <= n:
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itr1 = itr1.next
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i += 1
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# Now move the second iterator from head, till the first iterator reaches end.
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# If we need to find nth node from the last, we need to find l-nth node from the
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# beginning, where l is the length of the list.
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while itr1:
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itr1 = itr1.next
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itr2 = itr2.next
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return itr2.data
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def print_list(head):
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""" Prints a linked list.
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Params:
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head: head of the linked list
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"""
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while head:
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if not head.next:
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print(head.data)
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else:
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print(str(head.data) + " --> ", end = " ")
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head = head.next
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if __name__ == "__main__":
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root = Node(3)
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root.next = Node(2)
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root.next.next = Node(1)
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root.next.next.next = Node(5)
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root.next.next.next.next = Node(4)
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print_list(root)
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print("4th node from the last of the list is: ", nth_to_last(root, 4))

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