Posts tagged as “list”

花花酱 LeetCode 19. Remove Nth Node From End of List

By zxi on September 19, 2018

Problem

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

Solution 0: Cheating! store the nodes in an array

C++

// Author: Huahua

class Solution {

public:

ListNode *removeNthFromEnd(ListNode *head, int n) {

if (!head) return nullptr;

vector<ListNode *> nodes;

ListNode *cur = head;

while (cur) {

nodes.push_back(cur);

cur = cur->next;

}

if (n == nodes.size()) return head->next;

ListNode* nodes_to_remove = nodes[nodes.size()-n];

ListNode* parent = nodes[nodes.size() - n - 1];

parent->next = nodes_to_remove->next;

delete nodes_to_remove;

return head;

}

};

Solution 1: Two passes

Time complexity: O(L)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode* removeNthFromEnd(ListNode* head, int n) {

int l = 0;

ListNode* cur = head;

while (cur) {

++l;

cur = cur->next;

}

if (n == l) {

ListNode* ans = head->next;

delete head;

return ans;

}

l -= n;

cur = head;

while (--l) cur = cur->next;

ListNode* node = cur->next;

cur->next = node->next;

delete node;

return head;

}

};

Solution 2: Fast/Slow Pointers + Dummy Head / Prev

Fast pointer moves n steps first, and then slow pointer starts moving.

When fast pointer reaches tail, slow pointer is n-th node from the end.

Time complexity: O(L)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode* removeNthFromEnd(ListNode* head, int n) {

ListNode* fast = head;

for (int i = 0; i < n; ++i)

fast = fast->next;

ListNode dummy(0);

dummy.next = head;

ListNode* prev = &dummy;

while (fast) {

fast = fast->next;

prev = prev->next;

}

ListNode* node = prev->next;

prev->next = node->next;

delete node;

return dummy.next;

}

};

Java

// Author: Huahua

class Solution {

public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode fast = head;

while (n-- > 0) fast = fast.next;

ListNode prev = dummy;

while (fast != null) {

fast = fast.next;

prev = prev.next;

}

prev.next = prev.next.next;

return dummy.next;

}

Python3

# Author: Huahua

class Solution:

def removeNthFromEnd(self, head, n):

dummy = ListNode(0)

dummy.next = head

fast, prev = head, dummy

for _ in range(n):

fast = fast.next

while fast:

fast, prev = fast.next, prev.next

prev.next = prev.next.next

return dummy.next

花花酱 LeetCode 707. Design Linked List

By zxi on August 9, 2018

Problem

Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.

Implement these functions in your linked list class:

get(index) : Get the value of the index-th node in the linked list. If the index is invalid, return -1.
addAtHead(val) : Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
addAtTail(val) : Append a node of value val to the last element of the linked list.
addAtIndex(index, val) : Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
deleteAtIndex(index) : Delete the index-th node in the linked list, if the index is valid.

Example:

MyLinkedList linkedList = new MyLinkedList();

linkedList.addAtHead(1);

linkedList.addAtTail(3);

linkedList.addAtIndex(1, 2); // linked list becomes 1->2->3

linkedList.get(1); // returns 2

linkedList.deleteAtIndex(1); // now the linked list is 1->3

linkedList.get(1); // returns 3

Note:

All values will be in the range of [1, 1000].
The number of operations will be in the range of [1, 1000].
Please do not use the built-in LinkedList library.

Solution: Single linked list

Keep tracking head and tail of the list.

Time Complexity:

addAtHead, addAtTail O(1)

addAtIndex O(index)

deleteAtIndex O(index)

Space complexity: O(1)

C++

Tracking head/tail and size of the list.

// Author: Huahua

// Running time: 24 ms

class MyLinkedList {

public:

MyLinkedList(): head_(nullptr), tail_(nullptr), size_(0) {}

~MyLinkedList() {

Node* node = head_;

while (node) {

Node* cur = node;

node = node->next;

delete cur;

}

head_ = nullptr;

tail_ = nullptr;

}

int get(int index) {

if (index < 0 || index >= size_) return -1;

auto node = head_;

while (index--)

node = node->next;

return node->val;

}

void addAtHead(int val) {

head_ = new Node(val, head_);

if (size_++ == 0)

tail_ = head_;

}

void addAtTail(int val) {

auto node = new Node(val);

if (size_++ == 0) {

head_ = tail_ = node;

} else {

tail_->next = node;

tail_ = tail_->next;

}

void addAtIndex(int index, int val) {

if (index < 0 || index > size_) return;

if (index == 0) return addAtHead(val);

if (index == size_) return addAtTail(val);

Node dummy(0, head_);

Node* prev = &dummy;

while (index--) prev = prev->next;

prev->next = new Node(val, prev->next);

++size_;

}

void deleteAtIndex(int index) {

if (index < 0 || index >= size_) return;

Node dummy(0, head_);

Node* prev = &dummy;

for (int i = 0; i < index; ++i)

prev = prev->next;

Node* node_to_delete = prev->next;

prev->next = prev->next->next;

if (index == 0) head_ = prev->next;

if (index == size_ - 1) tail_ = prev;

delete node_to_delete;

--size_;

}

private:

struct Node {

int val;

Node* next;

Node(int _val): Node(_val, nullptr) {}

Node(int _val, Node* _next): val(_val), next(_next) {}

};

Node* head_;

Node* tail_;

int size_;

};

// Author: Huahua

// Running time: 28 ms

class MyLinkedList {

public:

MyLinkedList(): head_(nullptr), tail_(nullptr), dummy_(0), size_(0) {}

~MyLinkedList() {

Node* node = head_;

while (node) {

Node* cur = node;

node = node->next;

delete cur;

}

head_ = nullptr;

tail_ = nullptr;

}

int get(int index) {

if (index < 0 || index >= size_) return -1;

return getNode(index)->val;

}

void addAtHead(int val) {

head_ = new Node(val, head_);

if (size_++ == 0)

tail_ = head_;

}

void addAtTail(int val) {

auto node = new Node(val);

if (size_++ == 0) {

head_ = tail_ = node;

} else {

tail_->next = node;

tail_ = tail_->next;

}

void addAtIndex(int index, int val) {

if (index < 0 || index > size_) return;

if (index == 0) return addAtHead(val);

if (index == size_) return addAtTail(val);

Node* prev = getNode(index - 1);

prev->next = new Node(val, prev->next);

++size_;

}

void deleteAtIndex(int index) {

if (index < 0 || index >= size_) return;

Node* prev = getNode(index - 1);

Node* node_to_delete = prev->next;

prev->next = node_to_delete->next;

if (index == 0) head_ = prev->next;

if (index == size_ - 1) tail_ = prev;

delete node_to_delete;

--size_;

}

private:

struct Node {

int val;

Node* next;

Node(int _val): Node(_val, nullptr) {}

Node(int _val, Node* _next): val(_val), next(_next) {}

};

Node* head_; // Does not own

Node* tail_; // Does not own

Node dummy_;

int size_;

Node* getNode(int index) {

dummy_.next = head_;

Node* n = &dummy_;

for (int i = 0; i <= index; ++i)

n = n->next;

return n;

}

};

Python3

"""

Author: Huahua

Running time: 132 ms

"""

class Node:

def __init__(self, val, _next=None):

self.val = val

self.next = _next

class MyLinkedList:

def __init__(self):

self.head = self.tail = None

self.size = 0

def getNode(self, index):

n = Node(0, self.head)

for i in range(index + 1):

n = n.next

return n

def get(self, index):

if index < 0 or index >= self.size: return -1

return self.getNode(index).val

def addAtHead(self, val):

n = Node(val, self.head)

self.head = n

if self.size == 0:

self.tail = n

self.size += 1

def addAtTail(self, val):

n = Node(val)

if self.size == 0:

self.head = self.tail = n

else:

self.tail.next = n

self.tail = n

self.size += 1

def addAtIndex(self, index, val):

if index < 0 or index > self.size: return

if index == 0: return self.addAtHead(val)

if index == self.size: return self.addAtTail(val)

prev = self.getNode(index - 1)

n = Node(val, prev.next)

prev.next = n

self.size += 1

def deleteAtIndex(self, index):

if index < 0 or index >= self.size: return

prev = self.getNode(index - 1)

prev.next = prev.next.next

if index == 0: self.head = prev.next

if index == self.size - 1: self.tail = prev

self.size -= 1

Java

// Author: Huahua

// Running time: 74 ms

class MyLinkedList {

class Node {

public int val;

public Node next;

public Node(int val) { this.val = val; this.next = null; }

public Node(int val, Node next) { this.val = val; this.next = next; }

}

private Node head;

private Node tail;

private int size;

public MyLinkedList() {

this.head = this.tail = null;

this.size = 0;

}

private Node getNode(int index) {

Node n = new Node(0, this.head);

while (index-- >= 0) {

n = n.next;

}

return n;

}

public int get(int index) {

if (index < 0 || index >= size) return -1;

return getNode(index).val;

}

public void addAtHead(int val) {

this.head = new Node(val, this.head);

if (this.size++ == 0)

this.tail = this.head;

}

public void addAtTail(int val) {

Node n = new Node(val);

if (this.size++ == 0)

this.head = this.tail = n;

else

this.tail = this.tail.next = n;

}

public void addAtIndex(int index, int val) {

if (index < 0 || index > this.size) return;

if (index == 0) { this.addAtHead(val); return; }

if (index == size) { this.addAtTail(val); return; }

Node prev = this.getNode(index - 1);

prev.next = new Node(val, prev.next);

++this.size;

}

public void deleteAtIndex(int index) {

if (index < 0 || index >= this.size) return;

Node prev = this.getNode(index - 1);

prev.next = prev.next.next;

if (index == 0) this.head = prev.next;

if (index == this.size - 1) this.tail = prev;

--this.size;

}

花花酱 LeetCode 148. Sort List

By zxi on July 26, 2018

Problem

Sort a linked list in O(n log n) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

Solution: Merge Sort

Top-down (recursion)

Time complexity: O(nlogn)

Space complexity: O(logn)

C++

// Author: Huahua

// Running time: 32 ms

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* ListNode *next;

* ListNode(int x) : val(x), next(NULL) {}

* };

class Solution {

public:

ListNode* sortList(ListNode* head) {

// 0 or 1 element, we are done.

if (!head || !head->next) return head;

ListNode* slow = head;

ListNode* fast = head->next;

while (fast && fast->next) {

fast = fast->next->next;

slow = slow->next;

}

ListNode* mid = slow->next;

slow->next = nullptr; // Break the list.

return merge(sortList(head), sortList(mid));

}

private:

ListNode* merge(ListNode* l1, ListNode* l2) {

ListNode dummy(0);

ListNode* tail = &dummy;

while (l1 && l2) {

if (l1->val > l2->val) swap(l1, l2);

tail->next = l1;

l1 = l1->next;

tail = tail->next;

}

tail->next = l1 ? l1 : l2;

return dummy.next;

}

};

Java

// Author: Huahua

// Running time: 6 ms

class Solution {

public ListNode sortList(ListNode head) {

if (head == null || head.next == null) return head;

ListNode slow = head;

ListNode fast = head.next;

while (fast != null && fast.next != null) {

fast = fast.next.next;

slow = slow.next;

}

ListNode mid = slow.next;

slow.next = null;

return merge(sortList(head), sortList(mid));

}

private ListNode merge(ListNode l1, ListNode l2) {

ListNode dummy = new ListNode(0);

ListNode tail = dummy;

while (l1 != null && l2 != null) {

if (l1.val > l2.val) {

ListNode tmp = l1;

l1 = l2;

l2 = tmp;

}

tail.next = l1;

l1 = l1.next;

tail = tail.next;

}

tail.next = (l1 != null) ? l1 : l2;

return dummy.next;

}

Python3

# Author: Huahua, 232 ms

class Solution:

def sortList(self, head):

def merge(l1, l2):

dummy = ListNode(0)

tail = dummy

while l1 and l2:

if l1.val > l2.val: l1, l2 = l2, l1

tail.next = l1

l1 = l1.next

tail = tail.next

tail.next = l1 if l1 else l2

return dummy.next

if not head or not head.next: return head

slow = head

fast = head.next

while fast and fast.next:

fast = fast.next.next

slow = slow.next

mid = slow.next

slow.next = None

return merge(self.sortList(head), self.sortList(mid))

bottom up

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua

// Running time: 32 ms

class Solution {

public:

ListNode* sortList(ListNode* head) {

// 0 or 1 element, we are done.

if (!head || !head->next) return head;

int len = 1;

ListNode* cur = head;

while (cur = cur->next) ++len;

ListNode dummy(0);

dummy.next = head;

ListNode* l;

ListNode* r;

ListNode* tail;

for (int n = 1; n < len; n <<= 1) {

cur = dummy.next; // partial sorted head

tail = &dummy;

while (cur) {

l = cur;

r = split(l, n);

cur = split(r, n);

auto merged = merge(l, r);

tail->next = merged.first;

tail = merged.second;

}

return dummy.next;

}

private:

// Splits the list into two parts, first n element and the rest.

// Returns the head of the rest.

ListNode* split(ListNode* head, int n) {

while (--n && head)

head = head->next;

ListNode* rest = head ? head->next : nullptr;

if (head) head->next = nullptr;

return rest;

}

// Merges two lists, returns the head and tail of the merged list.

pair<ListNode*, ListNode*> merge(ListNode* l1, ListNode* l2) {

ListNode dummy(0);

ListNode* tail = &dummy;

while (l1 && l2) {

if (l1->val > l2->val) swap(l1, l2);

tail->next = l1;

l1 = l1->next;

tail = tail->next;

}

tail->next = l1 ? l1 : l2;

while (tail->next) tail = tail->next;

return {dummy.next, tail};

}

};

Posts tagged as “list”

花花酱 LeetCode 19. Remove Nth Node From End of List

Problem

Solution 0: Cheating! store the nodes in an array

C++

Solution 1: Two passes

C++

Solution 2: Fast/Slow Pointers + Dummy Head / Prev

C++

Java

Python3

花花酱 LeetCode 141. Linked List Cycle

Problem

Solution1: HashTable

Solution2: Fast + Slow pointers

花花酱 LeetCode 707. Design Linked List

Problem

Solution: Single linked list

花花酱 LeetCode 876. Middle of the Linked List

Problem

Solution: Slow + Fast Pointers

花花酱 LeetCode 148. Sort List

Problem

Solution: Merge Sort

C++

Java

Python3