Posts published in “List”

花花酱 LeetCode 817. Linked List Components

By zxi on April 14, 2018

Problem

题目大意：给你一个链表，再给你一些合法的节点，问你链表中有多少个连通分量（所有节点必须合法）。

https://leetcode.com/problems/linked-list-components/description/

We are given head, the head node of a linked list containing unique integer values.

We are also given the list G, a subset of the values in the linked list.

Return the number of connected components in G, where two values are connected if they appear consecutively in the linked list.

Example 1:

Input: 
head: 0->1->2->3
G = [0, 1, 3]
Output: 2
Explanation: 
0 and 1 are connected, so [0, 1] and [3] are the two connected components.

Example 2:

Input: 
head: 0->1->2->3->4
G = [0, 3, 1, 4]
Output: 2
Explanation: 
0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.

Note:

If N is the length of the linked list given by head, 1 <= N <= 10000.
The value of each node in the linked list will be in the range [0, N - 1].
1 <= G.length <= 10000.
G is a subset of all values in the linked list.

Solution1: Graph Traversal using DFS

// Author: Huahua

// Running time: 76 ms

class Solution {

public:

int numComponents(ListNode* head, vector<int>& G) {

unordered_set<int> f(G.begin(), G.end());

unordered_map<int, vector<int>> g;

int u = head->val;

while (head->next) {

head = head->next;

int v = head->val;

if (f.count(v) && f.count(u)) {

g[u].push_back(v);

g[v].push_back(u);

}

u = v;

}

int ans = 0;

unordered_set<int> visited;

for (int u : G) {

if (visited.count(u)) continue;

++ans;

dfs(u, g, visited);

}

return ans;

}

private:

void dfs(int cur, unordered_map<int, vector<int>>& g, unordered_set<int>& visited) {

if (visited.count(cur)) return;

visited.insert(cur);

for (const int next : g[cur])

dfs(next, g, visited);

}

};

Solution 2: Count tail node in sub graph

// Author: Huahua

// Running time: 35 ms

class Solution {

public:

int numComponents(ListNode* head, vector<int>& G) {

unordered_set<int> g(G.begin(), G.end());

int ans = 0;

while (head) {

if (g.count(head->val) && (!head->next || !g.count(head->next->val)))

++ans;

head = head->next;

}

return ans;

}

};

花花酱 LeetCode 328. Odd Even Linked List

By zxi on April 12, 2018

Problem

题目大意：给你一个链表，把所有奇数位置的节点串在一起，后面跟着所有串在一起的偶数位置的节点。

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on …

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Solution

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 21 ms

class Solution {

public:

ListNode* oddEvenList(ListNode* head) {

if (head == nullptr) return head;

ListNode dummy_odd(0);

ListNode dummy_even(0);

ListNode* prev_odd = &dummy_odd;

ListNode* prev_even = &dummy_even;

int index = 0;

while (head) {

auto next = head->next;

head->next = nullptr; // important

if (index++ & 1) {

prev_even->next = head;

prev_even = head;

} else {

prev_odd->next = head;

prev_odd = head;

}

head = next;

}

prev_odd->next = dummy_even.next;

return dummy_odd.next;

}

};

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

ListNode* oddEvenList(ListNode* head) {

if (head == nullptr) return head;

vector<ListNode> heads(2, ListNode(0));

vector<ListNode*> prevs{&heads[0], &heads[1]};

int index = 0;

while (head) {

auto next = head->next;

head->next = nullptr; // important

prevs[index]->next = head;

prevs[index] = head;

head = next;

index ^= 1;

}

prevs[0]->next = heads[1].next;

return heads[0].next;

}

};

Java

// Author: Huahua

// Running time: 1 ms

class Solution {

public ListNode oddEvenList(ListNode head) {

ListNode[] heads = new ListNode[]{new ListNode(0), new ListNode(0)};

ListNode[] prevs = new ListNode[]{heads[0], heads[1]};

int index = 0;

while (head != null) {

ListNode next = head.next;

head.next = null;

prevs[index].next = head;

prevs[index] = prevs[index].next;

head = next;

index ^= 1;

}

prevs[0].next = heads[1].next;

return heads[0].next;

}

Python3

"""

Author: Huahua

Running time: 56 ms

"""

class Solution:

def oddEvenList(self, head):

heads = [ListNode(0), ListNode(0)]

prevs = [heads[0], heads[1]]

index = 0

while head:

nxt = head.next

head.next = None

prevs[index].next = head

prevs[index] = prevs[index].next

head = nxt

index ^= 1

prevs[0].next = heads[1].next

return heads[0].next

花花酱 LeetCode 432. All O`one Data Structure

By zxi on April 6, 2018

Problem

题目大意：设计一种数据结构，支持inc/dec/getmaxkey/getminkey操作，必须都在O(1)时间内完成。

https://leetcode.com/problems/all-oone-data-structure/description/

Implement a data structure supporting the following operations:

Inc(Key) – Inserts a new key with value 1. Or increments an existing key by 1. Key is guaranteed to be a non-empty string.
Dec(Key) – If Key’s value is 1, remove it from the data structure. Otherwise decrements an existing key by 1. If the key does not exist, this function does nothing. Key is guaranteed to be a non-empty string.
GetMaxKey() – Returns one of the keys with maximal value. If no element exists, return an empty string "".
GetMinKey() – Returns one of the keys with minimal value. If no element exists, return an empty string "".

Challenge: Perform all these in O(1) time complexity.

Solution

Time complexity: O(1)

Space complexity: O(n), n = # of unique keys

// Author: Huahua

// Running time: 32 ms

class AllOne {

public:

/** Initialize your data structure here. */

AllOne() {}

/** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */

void inc(string key) {

auto it = m_.find(key);

if (it == m_.end()) {

if (l_.empty() || l_.front().value != 1)

l_.push_front({1, {key}});

else

l_.front().keys.insert(key);

m_[key] = l_.begin();

return;

}

auto lit = it->second;

auto nit = next(lit);

if (nit == l_.end() || nit->value != lit->value + 1)

nit = l_.insert(nit, {lit->value + 1, {}});

nit->keys.insert(key);

m_[key] = nit;

remove(lit, key);

}

/** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */

void dec(string key) {

auto it = m_.find(key);

if (it == m_.end()) return;

auto lit = it->second;

if (lit->value > 1) {

auto pit = prev(lit);

if (lit == l_.begin() || pit->value != lit->value - 1)

pit = l_.insert(lit, {lit->value - 1, {}});

pit->keys.insert(key);

m_[key] = pit;

} else {

// value == 1, remove from the data structure

m_.erase(key);

}

remove(lit, key);

}

/** Returns one of the keys with maximal value. */

string getMaxKey() {

return l_.empty() ? "" : *l_.back().keys.cbegin();

}

/** Returns one of the keys with Minimal value. */

string getMinKey() {

return l_.empty() ? "" : *l_.front().keys.cbegin();

}

private:

struct Node {

int value;

unordered_set<string> keys;

};

list<Node> l_;

unordered_map<string, list<Node>::iterator> m_;

// Remove from old node.

void remove(list<Node>::iterator it, const string& key) {

it->keys.erase(key);

if (it->keys.empty())

l_.erase(it);

}

};

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