LeetCode - Plus One

ou 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.

풀이(kotlin)

 - 맨 뒷자리에 1을 더해주고 10일경우는 0으로 바꾸고 그 앞자리에 1을 더해주면서 10이 아닐경우까지 반복한 뒤 배열을 리턴한다.

class Solution {
    fun plusOne(digits: IntArray): IntArray {
        val arr = IntArray(digits.size + 1)
        for (i in 1..arr.lastIndex) arr[i] = digits[i-1]
        arr[arr.lastIndex] = arr[arr.lastIndex] + 1
        for (i in arr.lastIndex downTo 1) {
            if (arr[i] == 10) {
                arr[i] = 0
                arr[i - 1] += 1
            } else break
        }
        return if (arr[0] == 0) {
            arr.filterIndexed { index, _ -> index > 0 }.toIntArray()
        } else arr
    }
}