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Commit a0b5e2f

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Merge pull request #12 from 149ps/LC-problems
LC Problems
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‎116. Populating Next Right Pointers in Each Node/116. Populating Next Right Pointers in Each Node.py‎

Lines changed: 4 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -28,16 +28,13 @@ def __init__(self, val, left, right, next):
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self.next = next
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"""
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class Solution:
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def connect(self, root: 'Node') -> 'Node':
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if not root:
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return None
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if root.left:
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root.left.next = root.right
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def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
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if not root: return None
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if root.right:
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if root.left:
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root.left.next = root.right
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if root.next:
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root.right.next = root.next.left
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else:
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root.right.next = None
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self.connect(root.left)
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self.connect(root.right)
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return root
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@@ -0,0 +1,48 @@
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"""
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Given a binary tree root and an integer target, delete all the leaf nodes with value target.
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Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).
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Example 1:
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Input: root = [1,2,3,2,null,2,4], target = 2
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Output: [1,null,3,null,4]
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Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left).
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After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).
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Example 2:
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Input: root = [1,3,3,3,2], target = 3
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Output: [1,3,null,null,2]
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Example 3:
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Input: root = [1,2,null,2,null,2], target = 2
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Output: [1]
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Explanation: Leaf nodes in green with value (target = 2) are removed at each step.
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Constraints:
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The number of nodes in the tree is in the range [1, 3000].
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1 <= Node.val, target <= 1000
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"""
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def removeLeafNodes(self, root: Optional[TreeNode], target: int) -> Optional[TreeNode]:
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if root.left: root.left = self.removeLeafNodes(root.left, target)
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if root.right: root.right = self.removeLeafNodes(root.right, target)
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return None if not root.left and not root.right and root.val == target else root
Lines changed: 79 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,79 @@
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"""
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You are given an array of integers nums and the head of a linked list. Return the head of the modified linked list after removing all nodes from the linked list that have a value that exists in nums.
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Example 1:
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Input: nums = [1,2,3], head = [1,2,3,4,5]
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Output: [4,5]
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Explanation:
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Remove the nodes with values 1, 2, and 3.
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Example 2:
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Input: nums = [1], head = [1,2,1,2,1,2]
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Output: [2,2,2]
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Explanation:
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Remove the nodes with value 1.
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Example 3:
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Input: nums = [5], head = [1,2,3,4]
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Output: [1,2,3,4]
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Explanation:
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No node has value 5.
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"""
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# Definition for singly-linked list.
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class ListNode:
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def __init__(self, val=0, next=None):
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self.val = val
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self.next = next
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class Solution:
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def modifiedList(self, nums, head):
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hset = set(nums)
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temp = ListNode(-100)
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temp.next = head
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result = temp
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while temp.next:
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if temp.next.val in hset:
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temp.next = temp.next.next
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else:
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temp = temp.next
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return result.next
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head = ListNode(1)
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head.next = ListNode(2)
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head.next.next = ListNode(3)
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head.next.next.next = ListNode(4)
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head.next.next.next.next = ListNode(5)
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# Sample input for nums
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nums = [1, 2, 3]
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# Create an instance of the Solution class
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solution = Solution()
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# Call the modifiedList method, passing both nums and head
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result = solution.modifiedList(nums, head)
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# Print the modified linked list
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while result:
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print(result.val)
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result = result.next
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"""
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Given the head of a singly linked list and an integer k, split the linked list into k consecutive linked list parts.
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The length of each part should be as equal as possible: no two parts should have a size differing by more than one. This may lead to some parts being null.
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The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later.
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Return an array of the k parts.
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Example 1:
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Input: head = [1,2,3], k = 5
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Output: [[1],[2],[3],[],[]]
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Explanation:
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The first element output[0] has output[0].val = 1, output[0].next = null.
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The last element output[4] is null, but its string representation as a ListNode is [].
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Example 2:
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Input: head = [1,2,3,4,5,6,7,8,9,10], k = 3
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Output: [[1,2,3,4],[5,6,7],[8,9,10]]
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Explanation:
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The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
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Constraints:
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The number of nodes in the list is in the range [0, 1000].
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0 <= Node.val <= 1000
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1 <= k <= 50
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"""
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def splitListToParts(self, head, k):
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count, temp = 0, head
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while temp:
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temp = temp.next
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count += 1
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chunk_size = count // k
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extra_part = count % k
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cur = head
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result = []
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for i in range(k):
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part_length = chunk_size + (1 if i < extra_part else 0)
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part_head = cur
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for j in range(part_length - 1):
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if cur:
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cur = cur.next
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if cur:
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next_part = cur.next
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cur.next = None
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cur = next_part
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result.append(part_head)
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return result

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