Plus One

最后更新于:2022-04-02 01:10:07

# Plus One ### Source - leetcode: [Plus One | LeetCode OJ](https://leetcode.com/problems/plus-one/) - lintcode: [(407) Plus One](http://www.lintcode.com/en/problem/plus-one/) ### 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. #### 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。 ### 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)O(n)O(n), 空间复杂度在最后一位有进位时恶化为 O(n)O(n)O(n), 当然也可以通过两次循环使得空间复杂度为 O(1)O(1)O(1). ### Reference - Soulmachine 的 leetcode 题解,将要加的数当做进位处理就是从这学到的。
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