[LeetCode#66]Plus One

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The problem:

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.

link: https://oj.leetcode.com/problems/plus-one/

My analysis:

This problem is a little tricky but meaningful!
The key point we should grasp tigitly is that we only add "1"(one digit) to the number.
The "1" have many hints in our algorithm design.
1. we can directly set the carry‘s initial value as 1. (add the program‘s elegance!)
2. we compute the carry for next iteration. thus, if we encounter the carry as 0, it means there would be no changes in the left partition(digits), we no longer need to proceed on.
3. if we scan the all digits of int[] digits, we encounter no 0. it means we have an carry exceeds the significant digit.
The tricky part is that, we could know the final value, cause we only add 1 to the original number. 9999 -> 10000.

My solution:

public class Solution {
    public int[] plusOne(int[] digits) {
        if (digits == null || digits.length == 0)
            return digits;
        
        int carry = 1; // assign 1 to carry directly
        int temp_digit;
        
        for (int i = digits.length - 1; i >= 0; i--) {
                temp_digit = (digits[i] + carry) % 10;
                carry = (digits[i] + carry) / 10;
                digits[i] = temp_digit;
                
                if (carry == 0)
                    return digits;
        }
        
        int[] ret = new int[digits.length + 1];
        ret[0] = 1;
        return ret;
    }
}

[LeetCode#66]Plus One

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原文地址：http://www.cnblogs.com/airwindow/p/4207982.html