[LeetCode - C] 66. Plus One

Plus One - LeetCode

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.

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Solution 1

 

일의 자리에 해당되는 부분 1을 증가시키고

 

그에 맞게 새로운 ret에 malloc 하여 리턴했다.

 

일의 자리에 해당되는 부분이 10이 되면

 

한 자릿수 늘어나니 그에 맞춰 처리했다.

 

int *plusOne(int *digits, int digitsSize, int *returnSize)
{
    int idx;
    int *ret;
    
    idx = digitsSize - 1;
    digits[idx]++;
    while (idx)
    {
        if (digits[idx] == 10)
        {
            digits[idx] = 0;
            digits[idx - 1]++;
        }
        idx--;
    }
    if (digits[idx] == 10)
    {
        ret = (int *)malloc(sizeof(int) * (digitsSize + 1));
        ret[0] = 1;
        digits[0] = 0;
        *returnSize = digitsSize + 1;
    }
    else
    {
        ret = (int *)malloc(sizeof(int) * digitsSize);
        *returnSize = digitsSize;
    }
    while (idx < digitsSize)
    {
        if (!digits[0])
            ret[idx + 1] = digits[idx];
        else
            ret[idx] = digits[idx];
        idx++;
    }
    return (ret);
}