Posts tagged as “list”

花花酱 LeetCode 160. Intersection of Two Linked Lists

By zxi on November 28, 2021

Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.

For example, the following two linked lists begin to intersect at node c1:

The test cases are generated such that there are no cycles anywhere in the entire linked structure.

Note that the linked lists must retain their original structure after the function returns.

Custom Judge:

The inputs to the judge are given as follows (your program is not given these inputs):

intersectVal – The value of the node where the intersection occurs. This is 0 if there is no intersected node.
listA – The first linked list.
listB – The second linked list.
skipA – The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node.
skipB – The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node.

The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be accepted.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Intersected at '8'
Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Intersected at '2'
Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: No intersection
Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Constraints:

The number of nodes of listA is in the m.
The number of nodes of listB is in the n.
0 <= m, n <= 3 * 10⁴
1 <= Node.val <= 10⁵
0 <= skipA <= m
0 <= skipB <= n
intersectVal is 0 if listA and listB do not intersect.
intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.

Follow up: Could you write a solution that runs in O(n) time and use only O(1) memory?

Solution 1: Two Passes by swapping heads

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

if (!headA || !headB) return nullptr;

ListNode* a = headA;

ListNode* b = headB;

while (a != b) {

a = a ? a->next : headB;

b = b ? b->next : headA;

}

return a;

}

};

花花酱 LeetCode 114. Flatten Binary Tree to Linked List

By zxi on November 27, 2021

Given the root of a binary tree, flatten the tree into a “linked list”:

The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Solution 1: Recursion

Time complexity: O(n)
Space complexity: O(|height|)

Python3

# Author: Huahua

class Solution:

def flatten(self, root: Optional[TreeNode]) -> None:

"""

Do not return anything, modify root in-place instead.

"""

def solve(root: Optional[TreeNode]):

if not root: return None, None

if not root.left and not root.right: return root, root

l_head = l_tail = r_head = r_tail = None

if root.left:

l_head, l_tail = solve(root.left)

if root.right:

r_head, r_tail = solve(root.right)

root.left = None

root.right = l_head or r_head

if l_tail:

l_tail.right = r_head

return root, r_tail or l_tail

return solve(root)[0]

Solution 2: Unfolding

Time complexity: O(n)
Space complexity: O(1)

Python3

# Author: Huahua

class Solution:

def flatten(self, root: Optional[TreeNode]) -> None:

while root:

if root.left:

prev = root.left

while prev.right: prev = prev.right

prev.right = root.right

root.right = root.left

root.left = None

root = root.right

花花酱 LeetCode 1823. Find the Winner of the Circular Game

By zxi on April 12, 2021

There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order. More formally, moving clockwise from the i^th friend brings you to the (i+1)^th friend for 1 <= i < n, and moving clockwise from the n^th friend brings you to the 1^st friend.

The rules of the game are as follows:

Start at the 1^st friend.
Count the next k friends in the clockwise direction including the friend you started at. The counting wraps around the circle and may count some friends more than once.
The last friend you counted leaves the circle and loses the game.
If there is still more than one friend in the circle, go back to step 2 starting from the friend immediately clockwise of the friend who just lost and repeat.
Else, the last friend in the circle wins the game.

Given the number of friends, n, and an integer k, return the winner of the game.

Example 1:

Input: n = 5, k = 2
Output: 3
Explanation: Here are the steps of the game:
1) Start at friend 1.
2) Count 2 friends clockwise, which are friends 1 and 2.
3) Friend 2 leaves the circle. Next start is friend 3.
4) Count 2 friends clockwise, which are friends 3 and 4.
5) Friend 4 leaves the circle. Next start is friend 5.
6) Count 2 friends clockwise, which are friends 5 and 1.
7) Friend 1 leaves the circle. Next start is friend 3.
8) Count 2 friends clockwise, which are friends 3 and 5.
9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.

Example 2:

Input: n = 6, k = 5
Output: 1
Explanation: The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1.

Constraints:

1 <= k <= n <= 500

Solution 1: Simulation w/ Queue / List

Time complexity: O(n*k)
Space complexity: O(n)

C++/Queue

// Author: Huahua, 56 ms, 24.3 MB

class Solution {

public:

int findTheWinner(int n, int k) {

queue<int> q;

for (int i = 1; i <= n; ++i)

q.push(i);

while (q.size() != 1) {

for (int i = 1; i < k; ++i) {

int x = q.front();

q.pop();

q.push(x);

}

q.pop();

}

return q.front();

}

};

C++/List

// Author: Huahua, 12 ms, 6.8 MB

class Solution {

public:

int findTheWinner(int n, int k) {

list<int> l;

for (int i = 1; i <= n; ++i)

l.push_back(i);

auto it = l.begin();

while (l.size() != 1) {

for (int i = 1; i < k; ++i)

if (++it == l.end())

it = l.begin();

it = l.erase(it);

if (it == l.end())

it = l.begin();

}

return l.front();

}

};

Posts tagged as “list”

花花酱 LeetCode 160. Intersection of Two Linked Lists

Solution 1: Two Passes by swapping heads

C++

花花酱 LeetCode 142. Linked List Cycle II

Solution 1: Hashtset

C++

Solution: Fast slow pointers

C++

花花酱 LeetCode 114. Flatten Binary Tree to Linked List

Solution 1: Recursion

Python3

Python3

花花酱 LeetCode 109. Convert Sorted List to Binary Search Tree

Solution 1: Recursion w/ Fast + Slow Pointers

C++

花花酱 LeetCode 1823. Find the Winner of the Circular Game

Solution 1: Simulation w/ Queue / List

C++/Queue

C++/List