Commit 23cdb66

committed

Add python solutions to leetcode 24 and 206

1 parent 7364940 commit 23cdb66Copy full SHA for 23cdb66

File tree

+75

-0

lines changed

+75

-0

lines changed

Lines changed: 45 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,45 @@`
	`1`	`+# Link: https://leetcode.com/problems/swap-nodes-in-pairs/`
	`2`	`+`
	`3`	`+######################`
	`4`	`+# ITERATIVE SOLUTION #`
	`5`	`+######################`
	`6`	`+`
	`7`	`+class IterativeSolution:`
	`8`	`+ def swapPairs(self, head: ListNode) -> ListNode:`
	`9`	`+ # EDGE CASE: Linked list has zero or one node`
	`10`	`+ if (head is None) or (head.next is None):`
	`11`	`+ return head`
	`12`	`+`
	`13`	`+ left_node, right_node = head, head.next`
	`14`	`+`
	`15`	`+ # Store the second node as the new head`
	`16`	`+ final_head = right_node`
	`17`	`+`
	`18`	`+ while left_node and right_node:`
	`19`	`+`
	`20`	`+ # Store reference to next left_node`
	`21`	`+ next_left = right_node.next`
	`22`	`+ if next_left:`
	`23`	`+ next_right = right_node.next.next`
	`24`	`+ else:`
	`25`	`+ next_right = None`
	`26`	`+`
	`27`	`+ # Swap the left_node and right_node`
	`28`	`+ right_node.next = left_node`
	`29`	`+ left_node.next = next_right`
	`30`	`+`
	`31`	`+ # If there is an odd number of nodes`
	`32`	`+ if (next_left) and (next_right is None):`
	`33`	`+ left_node.next = next_left`
	`34`	`+ break`
	`35`	`+`
	`36`	`+ # Move onto the next pair of nodes`
	`37`	`+ left_node = next_left`
	`38`	`+ right_node = next_right`
	`39`	`+`
	`40`	`+ return final_head`
	`41`	`+`
	`42`	`+`
	`43`	`+######################`
	`44`	`+# RECURSIVE SOLUTION #`
	`45`	`+######################`

Lines changed: 30 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,30 @@`
	`1`	`+# Link to problem: https://leetcode.com/problems/reverse-linked-list/`
	`2`	`+`
	`3`	`+# Definition for singly-linked list.`
	`4`	`+# class ListNode:`
	`5`	`+# def __init__(self, val=0, next=None):`
	`6`	`+# self.val = val`
	`7`	`+# self.next = next`
	`8`	`+class Solution:`
	`9`	`+`
	`10`	`+ def reverseList(self, head: ListNode) -> ListNode:`
	`11`	`+ # Base case: Last node or empty list`
	`12`	`+ if (head is None) or (head.next is None):`
	`13`	`+ return head`
	`14`	`+`
	`15`	`+ # Split up the linked list into two parts:`
	`16`	`+ # 1: first node`
	`17`	`+ # 2: rest of the list`
	`18`	`+`
	`19`	`+ # Let's reverse the rest of the list`
	`20`	`+ rest_list = self.reverseList(head.next);`
	`21`	`+`
	`22`	`+ # Now, let's connect the (reversed) rest of the list`
	`23`	`+ # to the first node. This means the first node now must`
	`24`	`+ # become the last node in the list`
	`25`	`+ head.next.next = head;`
	`26`	`+ head.next = None;`
	`27`	`+`
	`28`	`+ # Make sure you return the rest of the list!`
	`29`	`+ return rest_list;`
	`30`	`+`

Comments

(0)