LeetCode Problem 66 – Plus One: Description, Examples, Analysis, and Solutions in Java, C++, C, and Python

Recently a netizen shared a story about an interview where the interviewer turned out to be his ex‑girlfriend, leading to an awkward questioning session and a blocked phone number after the interview.

--------------下面是今天的算法题--------------

Problem Description : Given a non‑empty array of digits representing a non‑negative integer, add one to the integer. The most significant digit is at the start of the array, and each element stores a single digit. The integer does not have leading zeros except for zero itself.

Example 1 : Input: digits = [1,2,3] → Output: [1,2,4] (represents 123 → 124).

Example 2 : Input: digits = [4,3,2,1] → Output: [4,3,2,2] (represents 4321 → 4322).

Constraints : 1 ≤ digits.length ≤ 100, 0 ≤ digits[i] ≤ 9.

Problem Analysis : The task is to add one to a number represented by an array. If a digit is not 9, simply increment it and return. If a digit is 9, set it to 0 and continue the carry to the next higher digit. If all digits are 9, create a new array with length+1, set the first element to 1, and the rest to 0.

Java Solution :

public int[] plusOne(int[] digits) {
    int length = digits.length;
    // Traverse from the end
    for (int i = length - 1; i >= 0; i--) {
        if (digits[i] != 9) {
            // If current digit is not 9, add 1 and return
            digits[i]++;
            return digits;
        } else {
            // If current digit is 9, set to 0 and continue
            digits[i] = 0;
        }
    }
    // All digits were 9, need an extra digit
    int[] tmp = new int[length + 1];
    tmp[0] = 1;
    return tmp;
}

C++ Solution :

vector<int> plusOne(vector<int>& digits) {
    int length = digits.size();
    // Traverse from the end
    for (int i = length - 1; i >= 0; i--) {
        if (digits[i] != 9) {
            // If current digit is not 9, add 1 and return
            digits[i]++;
            return digits;
        } else {
            // If current digit is 9, set to 0 and continue
            digits[i] = 0;
        }
    }
    // All digits were 9, need an extra digit
    vector<int> tmp(length + 1);
    tmp[0] = 1;
    return tmp;
}

C Solution :

int* plusOne(int* digits, int digitsSize, int* returnSize) {
    // Traverse from the end
    for (int i = digitsSize - 1; i >= 0; i--) {
        if (digits[i] != 9) {
            // If current digit is not 9, add 1 and return
            digits[i]++;
            *returnSize = digitsSize;
            return digits;
        } else {
            // If current digit is 9, set to 0 and continue
            digits[i] = 0;
        }
    }
    // All digits were 9, need an extra digit
    int* tmp = (int*)malloc((digitsSize + 1) * sizeof(int));
    memset(tmp, 0, (digitsSize + 1) * sizeof(int));
    tmp[0] = 1;
    *returnSize = digitsSize + 1;
    return tmp;
}

Python Solution :

def plusOne(self, digits: List[int]) -> List[int]:
    length = len(digits)
    # Traverse from the end
    for i in range(length - 1, -1, -1):
        if digits[i] != 9:
            # If current digit is not 9, add 1 and return
            digits[i] += 1
            return digits
        else:
            # If current digit is 9, set to 0 and continue
            digits[i] = 0
    # All digits were 9, need an extra digit
    tmp = [0] * (length + 1)
    tmp[0] = 1
    return tmp