Plus One
Description
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.
Solution(javascript)
/**
* @param {number[]} digits
* @return {number[]}
*/
var plusOne = function(digits) {
// iterate through the array from last element to first
for (let i = digits.length - 1; i >= 0; i--) {
// increment the element
digits[i]++;
// if element is less than 10 and not the leftmost element, end iteration and return the array
if (digits[i] < 10 && digits[i] !== 0) return digits;
// element must be >10, so decrease by 10. No need to create a "carry" variable, because each iteration assumes the element needs to be incremented.
digits[i] = 0;
// if this is the leftmost element, we need to add one more element to the left to catch the carry
if (i === 0) {
digits.unshift(1)
return digits;
}
}
};