66. Plus One

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.
Example:
1.input [ 1, 1 ], return [ 1, 2 ]
2.input [ 9, 9 ], return [ 1, 0, 0]

Already Pass Solution

    public int[] PlusOne(int[] digits)
        {
            /*  -- Version One -- */
            int num = Int32.Parse(String.Join("", digits)) + 1;
            int length = digits.Length;
            digits[length - 1]++;
            for (int i = length - 1; i > 0; i--)
            {
                if (digits[i] > 9 && i != 0)
                {
                    digits[i - 1]++;
                    digits[i] -= 10;
                }

            }
            if (digits[0] > 9)
            {
                int[] result = new int[length + 1];
                result[0] = 1;
                for (int i = 0; i < result.Length; i++)
                {
                    result[i] %= 10;
                }
                return result;
            }
            else
            {
                return digits;
            }
        }

解题思路：
1.使用数组载体计算后判断长度是否有所增加，新建新的数组座位值返回

    public int[] PlusOne(int[] digits)
        {
            /*  -- Version Two -- */
            Stack<int> result = new Stack<int>();
            int index = digits.Length;
            digits[index-1]++;
            for(int i = index; i > 1; i--)
            {
                if (digits[i - 1] > 9)
                {
                    digits[i - 2]++;
                }
                result.Push(digits[i-1]%10);
            }
            result.Push(digits[0]%10);
            if (digits[0] > 9)
            {
                result.Push(1);
            }
            return result.ToArray();
        }

解题思路：
1.使用栈的数据结构辅助，减少代码量和循环遍历次数

待思考：
1.时间复杂度和空间复杂度降低