66. Plus One

Given a non-empty array of digits representing a non-negative integer, plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.

Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
//类似add binary : Time:  O(n), Space: O(1)
public int[] plusOne(int[] digits) {
        if (digits == null || digits.length == 0) {
            return new int[0];
        }
        
        int carrier = 1;
        
        for (int i = digits.length - 1; i >= 0; i--) {
            if (carrier == 0) {
                return digits;
            }//要把退出循环条件放在开始

            int sum = digits[i] + carrier;
            digits[i] = sum % 10;
            carrier = sum / 10;
        }
        
        int[] result = new int[digits.length + 1];
        result[0] = 1;
        
        return carrier == 0 ? digits : result;//注意出来一定要判定carrier是否为0,因为eg 399跳出循环
    }

 

posted @ 2018-10-17 14:06  一丝清风一抹红尘  阅读(128)  评论(0编辑  收藏  举报