66. Plus One (E)_66e.one-优快云博客

本文详细解析了一种算法问题：给定一个表示非负整数的数字数组，对其进行加一操作，并返回结果数组。通过实例展示了如何处理进位，确保算法正确性和效率。

Plus One (E)

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.

题意

将一个整数各位存放到数组中，输出加1后的整数(依然用数组表示)。

思路

注意进位即可。

代码实现

class Solution {
    public int[] plusOne(int[] digits) {
        int i = digits.length - 1;
        digits[i]++;
        int carry = digits[i] / 10;
        digits[i] = digits[i] % 10;
        // 需要进位则一直向前
        while (carry != 0 && i >= 1) {
            i--;
            digits[i] = digits[i] + carry;
            carry = digits[i] / 10;
            digits[i] = digits[i] % 10;
        }
        // 仍需要进位，说明需要新增一位数
        if (carry != 0) {
            int[] temp = new int[digits.length + 1];
            for (int j = 1; j < temp.length; j++) {
                temp[j] = digits[j - 1];
            }
            temp[0] = carry;
            digits = temp;
        }
        return digits;
    }
}