Plus One

Source

Given a non-negative number represented as an array of digits, plus one to the number.

The digits are stored such that the most significant digit is at the head of the list.

Example
Given [1,2,3] which represents 123, return [1,2,4].

Given [9,9,9] which represents 999, return [1,0,0,0].

题解

又是一道两个整数按数位相加的题，自后往前累加，处理下进位即可。这道题中是加1，其实还可以扩展至加2，加3等。

C++

class Solution {
public:
    /**
     * @param digits a number represented as an array of digits
     * @return the result
     */
    vector<int> plusOne(vector<int>& digits) {
        return plusN(digits, 1);
    }

    vector<int> plusN(vector<int>& digits, int n) {
        vector<int> result;
        int carry = n;
        for (int i = digits.size() - 1; i >= 0; i--) {
            result.insert(result.begin(), (digits[i] + carry) % 10);
            carry = (digits[i] + carry) / 10;
        }
        if (carry) result.insert(result.begin(), carry);
        return result;
    }
};

Java

public class Solution {
    /**
     * @param digits a number represented as an array of digits
     * @return the result
     */
    public int[] plusOne(int[] digits) {
        return plusDigit(digits, 1);
    }

    private int[] plusDigit(int[] digits, int digit) {
        if (digits == null || digits.length == 0) return null;

        // regard digit(0~9) as carry
        int carry = digit;
        int[] result = new int[digits.length];
        for (int i = digits.length - 1; i >= 0; i--) {
            result[i] = (digits[i] + carry) % 10;
            carry = (digits[i] + carry) / 10;
        }

        // carry == 1
        if (carry == 1) {
            int[] finalResult = new int[result.length + 1];
            finalResult[0] = 1;
            return finalResult;
        }

        return result;
    }
}

源码分析

源码中单独实现了加任何数(0~9)的私有方法，更为通用，对于末尾第一个数，可以将要加的数当做进位处理，这样就不必单独区分最后一位了，十分优雅！

复杂度分析

Java 中需要返回数组，而这个数组在处理之前是不知道大小的，故需要对最后一个进位单独处理。时间复杂度 O(n), 空间复杂度在最后一位有进位时恶化为 O(n), 当然也可以通过两次循环使得空间复杂度为 O(1).