https://leetcode.com/problems/count-and-say/description/
My GitHub: Count And Say

• The n^th level result is from counting each digit of the result of (n-1) ^th level.

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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

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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.

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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

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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};

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