Posts tagged as “list”

花花酱 LeetCode 2. Add Two Numbers

By zxi on July 25, 2018

Problem

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Solution: Simulation

Simulate the addition, draw two numbers (one each) from l1, l2, if list is empty draw 0.

Using a dummy head makes things easier.

Time complexity: O(max(l1,l2))

Space complexity: O(max(l1,l2))

C++

// Author: Huahua

// Running time: 28 ms

class Solution {

public:

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

ListNode dummy(0);

ListNode* tail = &dummy;

int sum = 0;

while (l1 || l2 || sum) {

sum += (l1 ? l1->val : 0) + (l2 ? l2->val : 0);

l1 = l1 ? l1->next : nullptr;

l2 = l2 ? l2->next : nullptr;

tail->next = new ListNode(sum % 10);

sum /= 10;

tail = tail->next;

}

return dummy.next;

}

};

Java

// Author: Huahua

// Running time: 30 ms

class Solution {

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

int sum = 0;

ListNode dummy = new ListNode(0);

ListNode tail = dummy;

while (l1 != null || l2 != null || sum != 0) {

if (l1 != null) {

sum += l1.val;

l1 = l1.next;

}

if (l2 != null) {

sum += l2.val;

l2 = l2.next;

}

tail.next = new ListNode(sum % 10);

sum /= 10;

tail = tail.next;

}

return dummy.next;

}

Python3

"""

Author: Huahua

Running time: 112 ms

"""

class Solution:

def addTwoNumbers(self, l1, l2):

s = 0

dummy = ListNode(0)

tail = dummy

while l1 or l2 or s > 0:

s += (l1.val if l1 else 0) + (l2.val if l2 else 0)

l1 = l1.next if l1 else None

l2 = l2.next if l2 else None

tail.next = ListNode(s % 10)

s //= 10

tail = tail.next

return dummy.next

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

Posts tagged as “list”

花花酱 LeetCode 2. Add Two Numbers

Problem

Solution: Simulation

C++

Java

Python3

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