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花花酱 LeetCode 445. Add Two Numbers II

Problem

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

Solution: Simulation

Using a stack to “reverse” the list. Simulate the addition digit by digit.

Time complexity: O(l1 + l2)

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

C++

Related Problems

花花酱 LeetCode 817. Linked List Components

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 head1 <= 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

Solution 2: Count tail node in sub graph

 

花花酱 LeetCode 328. Odd Even Linked List

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++

V2

Java

 

Python3

 

花花酱 LeetCode 432. All O`one Data Structure

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:

  1. 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.
  2. 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.
  3. GetMaxKey() – Returns one of the keys with maximal value. If no element exists, return an empty string "".
  4. 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

Related Problems

花花酱 LeetCode 382. Linked List Random Node

题目大意:写一个方法返回列表中的随机元素。

Problem:

https://leetcode.com/problems/linked-list-random-node/description/

Given a singly linked list, return a random node’s value from the linked list. Each node must have the same probability of being chosen.

Follow up:
What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without using extra space?

Example:

Solution:

C++

Time Complexity: O(n)

Space Complexity: O(1)