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花花酱 LeetCode 136. Single Number

By zxi on November 27, 2021

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

Input: nums = [2,2,1]
Output: 1

Example 2:

Input: nums = [4,1,2,1,2]
Output: 4

Example 3:

Input: nums = [1]
Output: 1

Constraints:

  • 1 <= nums.length <= 3 * 104
  • -3 * 104 <= nums[i] <= 3 * 104
  • Each element in the array appears twice except for one element which appears only once.

Solution: XOR

single_number ^ a ^ b ^ c ^ … ^ a ^ b ^ c … = single_number

Time complexity: O(n)
Space complexity: O(1)

C++

Related Problems

花花酱 LeetCode 119. Pascal’s Triangle II

By zxi on November 27, 2021

Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:

Input: rowIndex = 3
Output: [1,3,3,1]

Example 2:

Input: rowIndex = 0
Output: [1]

Example 3:

Input: rowIndex = 1
Output: [1,1]

Constraints:

  • 0 <= rowIndex <= 33

Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Solution: Remember last row

Time complexity: O(n2)
Space complexity: O(n)

C++

Related Problems

花花酱 LeetCode 118. Pascal’s Triangle

By zxi on November 27, 2021

Given an integer numRows, return the first numRows of Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:

Input: numRows = 5
Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]

Example 2:

Input: numRows = 1
Output: [[1]]

Constraints:

  • 1 <= numRows <= 30

Solution: Simulation

Time complexity: O(n2)
Space complexity: O(n2)

C++

Related Problems

花花酱 LeetCode 114. Flatten Binary Tree to Linked List

By zxi on November 27, 2021

Given the root of a binary tree, flatten the tree into a “linked list”:

  • The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
  • The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Solution 1: Recursion

Time complexity: O(n)
Space complexity: O(|height|)

Python3

Solution 2: Unfolding

Time complexity: O(n)
Space complexity: O(1)

Python3

花花酱 LeetCode 109. Convert Sorted List to Binary Search Tree

By zxi on November 27, 2021

Given the head of a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:

Input: head = []
Output: []

Example 3:

Input: head = [0]
Output: [0]

Example 4:

Input: head = [1,3]
Output: [3,1]

Constraints:

  • The number of nodes in head is in the range [0, 2 * 104].
  • -105 <= Node.val <= 105

Solution 1: Recursion w/ Fast + Slow Pointers

For each sublist, use fast/slow pointers to find the mid and build the tree.

Time complexity: O(nlogn)
Space complexity: O(logn)

C++