Plus One – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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 <= 1000 <= digits[i] <= 9digitsdoes not contain any leading0‘s.
C++ Plus One LeetCode Solution
class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
int n = digits.size() - 1;
for (int i = n; i >= 0; --i) { // traverse digits from the last element (least significant)
// since we begin with the last digit, increasing that digit by one
// results in overflow. Therefore, all elements PRIOR to digits[0]
// need to be considered since there may be additional nines between
// digits[0], ... , digits[n].
if (digits[i] == 9) {
digits[i] = 0;
} else { // current digit is not 9 so we can safely increment by one
digits[i] += 1;
return digits;
}
}
// if the program runs to this point, each 9 is now a 0.
// to get a correct solution, we need to add one more element with
// a value of zero AND set digits[0] to 1 (in the most significant position)
// to account for the carry digit.
digits.push_back(0);
digits[0] = 1;
return digits;
}
};
Java Plus One LeetCode Solution
public int[] plusOne(int[] digits) {
int n = digits.length;
for(int i=n-1; i>=0; i--) {
if(digits[i] < 9) {
digits[i]++;
return digits;
}
digits[i] = 0;
}
int[] newNumber = new int [n+1];
newNumber[0] = 1;
return newNumber;
}
Python 3 Plus One LeetCode Solution
def plusOne(digits):
num = 0
for i in range(len(digits)):
num += digits[i] * pow(10, (len(digits)-1-i))
return [int(i) for i in str(num+1)]
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