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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#
Easy

# Count and Say

# LeetCode – Plus One

# LeetCode- TWO SUM

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.

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};

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.