https://leetcode.com/problems/count-and-say/description/
My solution: Count And Say
- The nth level result is from counting each digit of the result of (n-1) th level.
Coding Interview
String to Integer (atoi)
https://leetcode.com/problems/string-to-integer-atoi/description/
github: String to Integer(atoi)
Checking point:
- null or empty string
- white spaces
- +/- sign
- calculate real value
- handle min & max
Using StringBuffer solve Reverse Integer problem
LeetCode: https://leetcode.com/problems/reverse-integer/description/
The below Solution 1 is the most popular solution we can find online.
But here is another way (Solution 2) to solve the problem with StringBuffer.
The Solution 2 is easier to understand and the code is cleaner.
leetcode- Rotate Image
Leetcode: https://leetcode.com/problems/rotate-image/description/
Given input matrix = [1,2,3], [4,5,6], [7,8,9] rotate the input matrix in-place such that it becomes: [7,4,1], [8,5,2], [9,6,3]
- (x, y), the affected point is
(y, n – x – 1), (n – x – 1, n – y – 1), (n – y – 1, x), “n” is the size of the matrix.
The rotate function=
(x, y) -> (y, n – x – 1) -> (n – x – 1, n – y – 1) -> (n – y – 1, x) -> (x, y). - For the i th line, we start from (i, i) ( !! in for loop j = i ) , and ends at (i, n – 2 – i).
- We only need process half of them. n/2
leetcode- Valid Sudoku
https://leetcode.com/problems/valid-sudoku/description/
In a 9 * 9 matrix, check if each numbers (0-9) doesn’t repeat in each row, col and 3 * 3 matrix.
- use a HashSet or array to check if the number is used.
- check 3*3 squart –> board [ i + k/3 ][ j + k%3 ] (i, j are 9*9 board index, k is the check board (0-9) index)
- if(digit > 9 || digit < 1 || check[digit -1]), check digit is valid first then check if the number is used
LeetCode – Plus One
Given a non-negative integer represented as a non-empty array of digits, plus one to the integer. You may assume the integer do not contain any leading zero, except the number 0 itself. The digits are stored such that the most significant digit is at the head of the list.
For example:
Q1.
int[] nums = {1,2,3,4};
int[] result = plusOne(nums);
the digits are in the result array are {1,2,3,5};
LeetCode- TWO SUM
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
Use a loop instead of two loops.
How to check if a string has all unique characters
GitHub: solution 1.1
- Check if the string is an ASCII string.
There are many different solutions, I practiced with a HashSet. Time complexity: O (n).
LeetCode – Two Sum II
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
CheckPoint: if (numbers != null && numbers.length > 1) {throw new IllegalArgumentException(“No solution!”);}
