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

easy

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

Problem

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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

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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

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• 1 <= digits.length <= 100
• 0 <= digits[i] <= 9
• digits does not contain any leading 0’s.

Solution

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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.

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[0] = 1;

    return digits;
  }
};