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
注意啦,这道题最要紧就是要读明白题,题中已经说了,数字会重复出现
可以参考快乐数的定义:
快乐数(happy
number)有以下的特性:在给定的进位制下,该数字所有数位(digits)的平方和,得到的新数再次求所有数位的平方和,如此重复进行,最终结果必为1。
不是快乐数的数称为不快乐数(unhappy
number),所有不快乐数的数位平方和计算,最後都会进入 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 的循环中。
我一开始就陷入到不是快乐数的循环中,因为会消耗掉栈内存啊!
看了大神们的答案后,发现原来不快乐数的原因是循环!有可能循环都是在10以上!
代码如下:
public
class Solution {
public boolean isHappy(int n) {
if(n==1)return true;
int fast,slow;
fast=slow=n;
do
{
slow=cal(slow);
fast=cal(fast);
fast=cal(fast);
}while(slow!=fast);
if(slow==1)return true;
else return false;
}
public int cal(int n)
{
int a=0;
while(n>0)
{
a+=Math.pow(n%10,2);
n/=10;
}
return a;
}
}
public boolean isHappy(int n) {
if(n==1)return true;
int fast,slow;
fast=slow=n;
do
{
slow=cal(slow);
fast=cal(fast);
fast=cal(fast);
}while(slow!=fast);
if(slow==1)return true;
else return false;
}
public int cal(int n)
{
int a=0;
while(n>0)
{
a+=Math.pow(n%10,2);
n/=10;
}
return a;
}
}