###### tags: `leetcode` `easy` `array` `math`
# [66. Plus One](https://leetcode.com/problems/plus-one/)
## Description
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.
## Examples
### 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=
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* plusOne(int* digits, int digitsSize, int* returnSize) {
int carry = 1;
// Sequentially checking each digit is carry or not
for (int i = digitsSize - 1; i >= 0; i--) {
digits[i] += carry;
if (digits[i] == 10) {
digits[i] = 0;
carry = 1;
continue;
}
carry = 0;
break;
}
// Using carry bit to assign correst return size
*returnSize = digitsSize + carry;
int* ans = malloc(*returnSize * sizeof(int));
// Assign number from digits with shifting of digits by carry
for (int i = *returnSize - 1; i > 0; i--)
ans[i] = digits[i - carry];
ans[0] = carry ? 1 : digits[0];
return ans;
}
```
## Complexity
|Space |Time |
|- |- |
|$O(n)$ |$O(n)$|
## Result
- Runtime: 2 ms, 76.18%
- Memory Usage: 5.8 MB, 30.18%