66. Plus One(Leetcode solution)

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.

SOLUTION:

Yours solution

class Solution {
public int[] plusOne(int[] digits) {
int i = digits.length — 1;//Getting the last index of digits

while(i >= 0 && digits[i] == 9){
i — ;
}

if(i == -1){//It means that all elements are 9 and we have to create a new array with length 1 more than the given array that is digits

int[] result = new int[digits.length + 1];
result[0] = 1;//if all elements are 9 in the digits array thanit means that after adding 1 the element at the 0 index will be 1 and all other are going to be 0

return result;

}

int[] result = new int[digits.length];
result[i] = digits[i] + 1;//idx where element doesn’t equal 9
for(int j = 0; j < i; j++){
result[j] = digits[j];
}

return result;
}
}