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 leading0
‘s.
Plus One Solutions
✅Time: O(n)
✅Space: O(1)
C++
class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
for (int i = digits.size() - 1; i >= 0; --i) {
if (digits[i] < 9) {
++digits[i];
return digits;
}
digits[i] = 0;
}
digits.insert(begin(digits), 1);
return digits;
}
};
Java
class Solution {
public int[] plusOne(int[] digits) {
for (int i = digits.length - 1; i >= 0; i--) {
if (digits[i] < 9) {
++digits[i];
return digits;
}
digits[i] = 0;
}
int[] ans = new int[digits.length + 1];
ans[0] = 1;
return ans;
}
}
Python
class Solution:
def plusOne(self, digits: List[int]) -> List[int]:
for i, d in reversed(list(enumerate(digits))):
if d < 9:
digits[i] += 1
return digits
digits[i] = 0
return [1] + digits
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