Plus One
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# LeetCode #66: Plus One (C/C++).

easy

source: https://leetcode.com/problems/plus-one/
C/C++ Solution to LeetCode problem 66. Plus One.

## Problem

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.

## Examples

### 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 = 
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

We iterate through the array from right to left.

• If the number is 9, change it to 0.
• If the number is less than 9 we add 1 and return the array.
• If we iterate through the whole array, we add a 0 to the end, and the first numbers becomes a 1.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 class Solution { public: vector<int> plusOne(vector<int>& digits) { int i = digits.size() - 1; while (i >= 0) { if (digits[i] == 9) digits[i] = 0; else { digits[i] += 1; return digits; } i--; } digits.push_back(0); digits = 1; return digits; } };