66. Plus One

⭐️

# 題目敘述

You are given a large integer represented as an integer array digits , where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0 's.

Increment the large integer by one and return the resulting array of digits.

# Example 1

Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].

# Example 2

Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].

# Example 3

Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].

# 解題思路

從陣列的最後面開始進行 +1 :

• 如果該數字為 0~8 ，將該數字 +1 ，回傳整個陣列。
• 如果該數字為 9 ，將該數字記為 0 ，繼續向前一位的陣列數字 +1 。

如果一直加到最前面的陣列數字都還是 9 ，我們就需要創建一個比原先傳入陣列還要大一位的陣列，將首位數字設為 1 ，其餘為 0 。

# Solution

class Solution {

    public int[] plusOne(int[] digits) {

        for(int i = digits.length - 1; i >= 0; i--) {

            if (digits[i] < 9) {

                digits[i]++;

                return digits;

            }

            digits[i] = 0;

        }

        int[] newNumber = new int[digits.length + 1];

        newNumber[0] = 1;

        return newNumber;

    }

}