66. Plus One
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].
Solution:
class Solution {
public int[] plusOne(int[] digits) {
// base case
if (digits == null || digits.length == 0){
return new int[]{1};
}
if (digits.length == 1 && digits[0] <= 8){
digits[0] = digits[0] + 1;
return digits;
}
if (digits.length == 1 && digits[0] == 9){
return new int[]{1,0};
}
digits[digits.length - 1] = digits[digits.length -1] + 1;
int[] result = new int[digits.length + 1];
int carry = 0;
for (int i = digits.length - 1; i >= 0; i--){
digits[i] = digits[i] + carry;
carry = digits[i] / 10;
digits[i] = digits[i] % 10;
result[i+1] = digits[i];
}
result[0] = carry;
if (result[0] != 0){
return result;
}else{
return digits;
}
}
}
// TC: O(n)
// SC: O(n)