66. Plus One

Problem

Tags: Array, Math

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.

Code

C

// 66. Plus One (3/8/54140)
// Runtime: 0 ms (94.74%) Memory: 6.07 MB (22.23%) 

int* plusOne (int* digits, int size, int* new_size) {
    int* new_digits = calloc(size + 1, sizeof(int));
    new_digits[size] = 1;
    
    for (int i = size - 1; i >= 0; i--) {
        new_digits[i + 1] = digits[i] + new_digits[i + 1];
        new_digits[i] = new_digits[i + 1] / 10;
        new_digits[i + 1] %= 10;
    }
    
    if (new_digits[0]) {
        *new_size = size + 1;
        return new_digits;
    } else {
        *new_size = size;
        return new_digits + 1;
    }
}