【Plus One】-（Java）

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].

Constraints:

• 1 <= digits.length <= 100
• 0 <= digits[i] <= 9
• digits does not contain any leading 0’s.

completion:

class Solution {
    /**
     * 分析：1.从最后1位开始+1，进位后继续+1，不进位后中止，原样输出（非全9的情况）
     * 		2.如果第1位为9，并且进位，则新声明一个数组，第一位为1，其他为0（9、99、999、9999,只有全9的情况下，+1才能改变位数）
     * @param digits
     * @return
     */
    public int[] plusOne(int[] digits) {
        int j = digits.length - 1;
        for (int i = j; i >= 0; i--) {
            if (digits[i] < 9) {
                digits[i]++;
                return digits;
            }
            digits[i] = 0;
        }
        //int默认值为0
        int[] number = new int[digits.length + 1];
        number[0] = 1;
        return number;
    }
}