Problem
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
ExampleGiven num = 38.
The process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return 2.
Could you do it without any loop/recursion in O(1) runtime?
Solutionpublic class Solution { /* * @param num: a non-negative integer * @return: one digit */ public int addDigits(int num) { // write your code here if (num < 10) return num; int sum = 0; while (num != 0) { sum += (num%10); num /= 10; } return addDigits(sum); } }
public class Solution { /* * @param num: a non-negative integer * @return: one digit */ public int addDigits(int num) { // write your code here while (num >= 10) { String cur = Integer.toString(num); int sum = 0; for (int i = 0; i < cur.length(); i++) { sum += Character.getNumericValue(cur.charAt(i)); } num = sum; } return num; } }
文章版權歸作者所有,未經允許請勿轉載,若此文章存在違規行為,您可以聯系管理員刪除。
轉載請注明本文地址:http://specialneedsforspecialkids.com/yun/68260.html
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...
Problem Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2. Notice The length of both num1 and num2 is < 5100.Both num1 and num2 contains only digits ...
摘要:題意為取刪去個字符后最小的值,仍以返回。所以無論刪除個元素之后的元素放入順序如何,此時棧內元素從底到頂的排列一定是滿足條件的最小值。這種情況下,就要從堆棧頂部刪除剩下的兩個元素和然后,將棧內的元素放入,并將頂部的全部去掉,然后以返回。 Problem Given string A representative a positive integer which has N digits,...
Problem You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1s digit is at the head of the list. Write a f...
摘要:只有當位數時,才打印數字。首先分析邊界,應該,然后用存最高位。用函數對進行遞歸運算,同時更新結果數組。更新的過程歸納一下,首先,計算最高位存入,然后,用到倍的和之前里已經存入的所有的數個循環相加,再存入,更新,計算更高位直到等于 Problem Print numbers from 1 to the largest number with N digits by recursion. ...
閱讀 1399·2021-11-08 13:14
閱讀 754·2021-09-23 11:31
閱讀 1046·2021-07-29 13:48
閱讀 2786·2019-08-29 12:29
閱讀 3380·2019-08-29 11:24
閱讀 1905·2019-08-26 12:02
閱讀 3695·2019-08-26 10:34
閱讀 3441·2019-08-23 17:07