标签：
Write an algorithm to determine if a number is "happy". A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers. Example: 19 is a happy number 12 + 92 = 82 82 + 22 = 68 62 + 82 = 100 12 + 02 + 02 = 1
First Time: Not good, when I try to get to the most significant bit, I define current bit called cur, and cur *= 10 each time. This might gives rise to overflow in line 12 when cur * 10. To overcome this I have to define cur to be a long.
The best way to get each digit is to use n%10, and n /=10 each loop. Dividing will avoid unnecessary overflow.
BTW, no need to define an arraylist to store all the digits.
1 public class Solution { 2 public boolean isHappy(int n) { 3 if (n <= 0) return false; 4 if (n == 1) return true; 5 HashSet<Integer> set = new HashSet<Integer>(); 6 while(!set.contains(n)) { 7 set.add(n); 8 long cur = 1; 9 ArrayList<Integer> digits = new ArrayList<Integer>(); 10 int sum = 0; 11 while (n / cur >= 1) { 12 digits.add((int)(n % (cur*10) / cur)); 13 cur *= 10; 14 } 15 for (int digit : digits) { 16 sum += (int)Math.pow(digit, 2); 17 } 18 n = sum; 19 if (n == 1) return true; 20 } 21 return false; 22 } 23 }
About syntax, line 16 it is correct to write it as: sum += Math.pow(digit, 2),
but it is incorrect as: sum = sum + Math.pow(digit, 2), should write as sum = sum + (int)Math.pow(digit, 2)
So better cast pow into integer before use it.
Second Time: Better
The best way to get each digit is to use n%10, and n /=10 each loop.
1 public class Solution { 2 public boolean isHappy(int n) { 3 if (n <= 0) return false; 4 if (n == 1) return true; 5 HashSet<Integer> set = new HashSet<Integer>(); 6 while(!set.contains(n)) { 7 set.add(n); 8 int sum = 0; 9 while (n > 0) { 10 sum += (int)Math.pow(n%10, 2); 11 n /= 10; 12 } 13 n = sum; 14 if (n == 1) return true; 15 } 16 return false; 17 } 18 }
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原文地址：http://www.cnblogs.com/EdwardLiu/p/5046973.html
