66. Plus One

🟩 Easy

You are given a large integer represented as an integer array digits, where each digits[i] is the i^th 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.

Solution

My Solution

func plusOne(digits []int) []int {
    n := len(digits) - 1

    for i := n; i >= 0; i-- {
        if digits[i] == 9 {
            digits[i] = 0
            if i == 0 {
                digits = append([]int{1}, digits...)
            }
        } else {
            digits[i]++
            break
        }

    }

    return digits
}

Optimal Solution

func plusOne(digits []int) []int {
    n := len(digits)
    
    for i := n - 1; i >= 0; i-- {
        if digits[i] < 9 {
            digits[i]++
            return digits
        }
        digits[i] = 0
    }
    
    // If all digits are 9, then we need to add a new digit
    return append([]int{1}, digits...)
}

result

Leetcode: link