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A6:Add Digits
题目描述:
Given a non-negative integer num
, repeatedly add all its digits until the result has only one digit.
样例
Given num
= 38.
The process is like: 3 + 8 = 11
, 1 + 1
= 2
. Since 2
has only one digit, return 2
.
解答:
1 public class Solution { 2 /** 3 * @param num a non-negative integer 4 * @return one digit 5 */ 6 public int addDigits(int num) { 7 // Write your code here 8 while(num / 10 > 0){ 9 int[] tmp = getNum(num); 10 num = 0; 11 for(int i=0; i<tmp.length; i++){ 12 num = num + tmp[i]; 13 } 14 } 15 return num; 16 } 17 18 private int[] getNum(int num){ 19 int length = Integer.toString(num).length(); 20 int[] tmp = new int[length]; 21 int i = 0; 22 while (num / 10 > 0){ 23 24 tmp[i] = num % 10; 25 num = num / 10 ; 26 i++; 27 } 28 tmp[length-1] = num; 29 return tmp; 30 } 31 32 }
A6:Add Digits
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