GuilinDev

Lc0066

05 August 2008

66 - Plus One

原题概述

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:

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

Example 2:

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Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.

题意和分析

这道题目很简单，就是需要考虑数组中最后一位加1的时候进位的问题。

Time: O(n)

代码

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class Solution {
    public int[] plusOne(int[] digits) {
        int carry = 1;
        
        //如果carry==0，没有往前进位的情况，就终止循环，不再加了
        for (int i = digits.length - 1; i >= 0 && carry > 0; i--) {
            int sum = digits[i] + carry;
            digits[i] = sum % 10;
            carry = sum / 10;
        }
        
        //处理数组中最后一次加法（第0位）可能往前进位的情况
        if (carry == 0) {//没有进位了，直接返回
            return digits;
        } else { //有进位，新建一个长度多一位的array
            int[] newDigits = new int[digits.length + 1];
            newDigits[0] = 1;
            for (int j = 1; j <= digits.length; j++) {
                newDigits[j] = digits[j - 1];
            }
            return newDigits;
        }
    }
}
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