Happy Number

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

1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 =1

分析:
这题的关键是如何退出循环如果那个数不是hapy number. 这里用了一个方法,就是用hashset保存得到过的数,如果出现重复,那就不是Happy number.

 1 public class Solution {
 2     /**
 3      * @param n an integer
 4      * @return true if this is a happy number or false
 5      */
 6     public boolean isHappy(int n) {
 7         if (n < 1) return false;
 8         
 9         Set<Integer> hashSet = new HashSet<Integer>();
10         hashSet.add(n);
11         
12         while (n != 1) {
13             n = getNewNumber(n);
14             if (hashSet.contains(n)) {
15                 return false;
16             }
17             hashSet.add(n);
18         }
19         return true;
20     }
21     
22     public int getNewNumber(int n) {
23         int total = 0;
24         while (n != 0) {
25             total += (n % 10) * (n % 10);
26             n = n / 10;
27         }
28         return total;
29     }
30 }

 

 1 public class Solution {
 2     int digitSquareSum(int n) {
 3         int sum = 0, tmp;
 4         while (n > 0) {
 5             tmp = n % 10;
 6             sum += tmp * tmp;
 7             n /= 10;
 8         }
 9         return sum;
10     }
11 
12     boolean isHappy(int n) {
13         int slow, fast;
14         slow = fast = n;
15         do {
16             slow = digitSquareSum(slow);
17             fast = digitSquareSum(fast);
18             fast = digitSquareSum(fast);
19         } while(slow != fast);
20         if (slow == 1) return true;
21         return false;
22     }
23 }

 

 
posted @ 2016-07-12 06:49  北叶青藤  阅读(426)  评论(0编辑  收藏  举报