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花花酱 LeetCode 725. Split Linked List in Parts

By zxi on November 13, 2017

Problem:

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list “parts”.

The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.

Return a List of ListNode’s representing the linked list parts that are formed.

Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]

Example 1:

Input:

root = [1, 2, 3], k = 5

Output: [[1],[2],[3],[],[]]

Explanation:

The input and each element of the output are ListNodes, not arrays.

For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.

The first element output[0] has output[0].val = 1, output[0].next = null.

The last element output[4] is null, but it's string representation as a ListNode is [].

Example 2:

Input:

root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3

Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]

Explanation:

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.

Note:

The length of root will be in the range [0, 1000].
Each value of a node in the input will be an integer in the range [0, 999].
k will be an integer in the range [1, 50].

Idea:

List + Simulation

Solution:

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

vector<ListNode*> splitListToParts(ListNode* root, int k) {

int len = 0;

for (ListNode* head = root; head; head = head->next) ++len;

vector<ListNode*> ans(k, nullptr);

int l = len / k;

int r = len % k;

ListNode* head = root;

ListNode* prev = nullptr;

for (int i = 0; i < k; ++i, --r) {

ans[i] = head;

for (int j = 0; j < l + (r > 0); ++j) {

prev = head;

head = head->next;

}

if (prev) prev->next = nullptr;

}

return ans;

}

};

Java

// Author: Huahua

// Runtime: 4 ms

class Solution {

public ListNode[] splitListToParts(ListNode root, int k) {

ListNode[] ans = new ListNode[k];

int len = 0;

for (ListNode head = root; head != null; head = head.next) ++len;

int l = len / k;

int r = len % k;

ListNode prev = null;

ListNode head = root;

for (int i = 0; i < k; ++i, --r) {

ans[i] = head;

int part_len = l + ((r > 0) ? 1 : 0);

for (int j = 0; j < part_len; ++j) {

prev = head;

head = head.next;

}

if (prev != null) prev.next = null;

}

return ans;

}

Python

"""

Author: Huahua

Runtime: 79 ms

"""

class Solution:

def splitListToParts(self, root, k):

total_len = 0

head = root

while head:

total_len += 1

head = head.next

ans = [None for _ in range(k)]

l, r = total_len // k, total_len % k

prev, head = None, root

for i in range(k):

ans[i] = head

for j in range(l + (1 if r > 0 else 0)):

prev, head = head, head.next

if prev: prev.next = None

r -= 1

return ans

花花酱 LeetCode 460. LFU Cache

By zxi on September 14, 2017

Problem:

Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the following operations: get and put.

get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) – Set or insert the value if the key is not already present. When the cache reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LFUCache cache = new LFUCache( 2 /* capacity */ );

cache.put(1, 1);

cache.put(2, 2);

cache.get(1); // returns 1

cache.put(3, 3); // evicts key 2

cache.get(2); // returns -1 (not found)

cache.get(3); // returns 3.

cache.put(4, 4); // evicts key 1.

cache.get(1); // returns -1 (not found)

cache.get(3); // returns 3

cache.get(4); // returns 4

Idea:

Hashtable + balanced search tree

Hashtable + double linked list

Solution 1: O(logc) c is the capacity

// Author: Huahua

// Running time: 99 ms

struct CacheNode {

int key;

int value;

int freq;

long tick;

// Defines the order, smaller one will be evicted first

bool operator <(const CacheNode& rhs) const {

if (freq < rhs.freq) return true;

if (freq == rhs.freq) return tick < rhs.tick;

return false;

}

};

class LFUCache {

public:

LFUCache(int capacity):capacity_(capacity), tick_(0) {}

int get(int key) {

auto it = m_.find(key);

if (it == m_.cend()) return -1;

int value = it->second.value;

touch(it->second);

return value;

}

void put(int key, int value) {

if (capacity_ == 0) return;

auto it = m_.find(key);

if (it != m_.cend()) {

// Key exists, update value and touch

it->second.value = value;

touch(it->second);

return;

}

if (m_.size() == capacity_) {

// Remove the first node in cache

const CacheNode& node = *cache_.cbegin();

m_.erase(node.key);

cache_.erase(node);

}

CacheNode node{key, value, 1, ++tick_};

m_[node.key] = node;

cache_.insert(node);

}

private:

void touch(CacheNode& node) {

cache_.erase(node); // log(capacity)

++node.freq;

node.tick = ++tick_;

cache_.insert(node); // log(capacity)

}

long tick_;

int capacity_;

unordered_map<int, CacheNode> m_;

set<CacheNode> cache_;

};

Solution 2: O(1)

// Author: Huahua

// Running time: 89 ms (beats 96.3%)

struct CacheNode {

int key;

int value;

int freq;

// pointer to the node in the list

list<int>::const_iterator it;

};

class LFUCache {

public:

LFUCache(int capacity): capacity_(capacity), min_freq_(0) {}

int get(int key) {

auto it = n_.find(key);

if (it == n_.cend()) return -1;

touch(it->second);

return it->second.value;

}

void put(int key, int value) {

if (capacity_ == 0) return;

auto it = n_.find(key);

if (it != n_.cend()) {

// Key already exists, update the value and touch it

it->second.value = value;

touch(it->second);

return;

}

if (n_.size() == capacity_) {

// No capacity, need to remove one entry that

// 1. has the lowest freq

// 2. least recently used if there are multiple ones

// Step 1: remove the element from min_freq_ list

const int key_to_evict = l_[min_freq_].back();

l_[min_freq_].pop_back();

// Step 2: remove the key from cache

n_.erase(key_to_evict);

}

// We know new item has freq of 1, thus set min_freq to 1

const int freq = 1;

min_freq_ = freq;

// Add the key to the freq list

l_[freq].push_front(key);

// Create a new node

n_[key] = {key, value, freq, l_[freq].cbegin()};

}

private:

void touch(CacheNode& node) {

// Step 1: update the frequency

const int prev_freq = node.freq;

const int freq = ++(node.freq);

// Step 2: remove the entry from old freq list

l_[prev_freq].erase(node.it);

if (l_[prev_freq].empty() && prev_freq == min_freq_) {

// Delete the list

l_.erase(prev_freq);

// Increase the min freq

++min_freq_;

}

// Step 4: insert the key into the front of the new freq list

l_[freq].push_front(node.key);

// Step 5: update the pointer

node.it = l_[freq].cbegin();

}

int capacity_;

int min_freq_;

// key -> CacheNode

unordered_map<int, CacheNode> n_;

// freq -> keys with freq

unordered_map<int, list<int>> l_;

};

花花酱 LeetCode 21: Merge Two Sorted Lists

By zxi on September 3, 2017

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Solution 1: Iterative O(n)

// Author: Huahua

class Solution {

public:

Solution 1: Iterative

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

ListNode dummy(0);

ListNode* tail=&dummy;

while(l1 && l2) {

if(l1->val < l2->val) {

tail->next=l1;

l1=l1->next;

}else{

tail->next=l2;

l2=l2->next;

}

tail=tail->next;

}

if(l1) tail->next = l1;

if(l2) tail->next = l2;

return dummy.next;

}

};

Solution 2: Recursive O(n)

// Author: Huahua

class Solution {

public:

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

// If one of the list is emptry, return the other one.

if(!l1 || !l2) return l1 ? l1 : l2;

// The smaller one becomes the head.

if(l1->val < l2->val) {

l1->next = mergeTwoLists(l1->next, l2);

return l1;

} else {

l2->next = mergeTwoLists(l1, l2->next);

return l2;

}

};

Posts tagged as “list”

花花酱 LeetCode 725. Split Linked List in Parts

花花酱 LeetCode 460. LFU Cache

花花酱 LeetCode 21: Merge Two Sorted Lists