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 = [0]
**Output:** [1]
**Explanation:** The array represents the integer 0.
Incrementing by one gives 0 + 1 = 1.
Thus, the result should be [1].
Example 4:
**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.
ac
classSolution {publicint[] plusOne(int[] digits) {// edge casesif (digits ==null||digits.length==0) returnnewint[0];// iterateint n =digits.length;int carry = (digits[n-1] +1) /10; digits[n-1] = (digits[n-1] +1) %10;for (int i = n -2; i >=0&& carry >0; i--) {int tmp = carry; carry = (tmp + digits[i]) /10; digits[i] = (tmp + digits[i]) %10; }// more digits than originalint[] res =newint[n+1];if (carry >0) { res[0] = carry;for (int i =0; i < n; i++) { res[i+1] = digits[i]; } } else { res = digits; }return res; }}/*carry*/