zoukankan      html  css  js  c++  java
  • [LeetCode#66]Plus One

    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;
        }
    }
  • 相关阅读:
    高效能人士懂得忽视,知道怎样说“不”
    CheckBoxPreference组件
    SQL基础--> 约束(CONSTRAINT)
    html5--6-7 CSS选择器4
    html5--6-6 CSS选择器3
    html5--6-5 CSS选择器2
    html5--6-4 CSS选择器
    html5--6-3 CSS语法2
    html5--6-2 CSS语法
    html5--6-1 引入外部样式表
  • 原文地址:https://www.cnblogs.com/airwindow/p/4207982.html
Copyright © 2011-2022 走看看