Plus One Problem

Description

LeetCode Problem 66.

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:

``````1
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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:

``````1
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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:

``````1
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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:

``````1
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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.

Sample C++ Code

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class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
int n = digits.size();
vector<int> ans;
int c = 0, sum;
for (int i = n-1; i >= 0; i --) {
if (i == n-1)
sum = digits[i] + 1;
else
sum = digits[i] + c;
c = sum / 10;
ans.push_back(sum % 10);
}
if (c != 0)
ans.push_back(c);
reverse(ans.begin(), ans.end());
return ans;
}
};
``````