# 66 Plus One

Edge Case 很重要，這題就考 [9, 9, 9需要進位時怎麼辦]

LeetCode

Given a non-empty array of digits representing a non-negative integer, 
plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, 
and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

input: 給一個正整數組成且排序好的陣列，可能有 0
output: 最後一個數子 + 1

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.


Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.

/**
 * @param {number[]} digits
 * @return {number[]}
 */
var plusOne = function(digits) {
    
};

Edge

最後一位數是 9 就需要進位，所以可能有 input: 999 ，output: 1000 的狀況

怎麼解

• 若最後一個數字是 0 - 8 那就直接 + 1

• 若最後一位是 9 那就自己變 0, 往前檢查 + 1

• 會先設 pointer = digits.length - 1

var plusOne = function(digits) {
  // think if digits.length < 9 then ++
  // else digits.length = 0 and check previous one
  // loop
  let pointer = digits.length - 1;
  
  while( digits[pointer] == 9){
      digits[pointer] = 0
      pointer --;
  }

  digits[pointer] ++;
  
  // if all 9 eg. 9, 9, 9
  if(pointer < 0){
    digits.unshift(1)
  }
  
  return digits
  
};