202 Happy Number 快乐数
Description:
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:
Input: 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
题目描述:
编写一个算法来判断一个数是不是“快乐数”。
一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和，然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那么这个数就是快乐数。
示例：
输入: 19
输出: true
解释:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
思路:
快乐数定义参考快乐数 Happy number
也可以自己打印循环中的数, 验证一下
不快乐数一定会进入
4 -> 16 -> 37 -> 58 -> 89 -> 145 -> 42 -> 20 -> 4
循环
- 每次计算结果加入哈希表, 判断哈希表中是否存在, 或计算结果是否为1
- 判断计算结果为1或4
时间复杂度O(1), 空间复杂度O(1)
因为 n最大为 2 ^ 31 - 1, 在 4~5次循环后就会收敛到 1000内, 最多再计算 1000次就一定能跳出循环
采用 set空间复杂度会增加到 n
代码:
C++:
class Solution
{
public:
bool isHappy(int n)
{
while (n != 1 and n != 4)
{
int temp = 0;
while (n)
{
temp += pow(n % 10, 2);
n /= 10;
}
n = temp;
}
return n == 1;
}
};
Java:
class Solution {
public boolean isHappy(int n) {
Set<Integer> set = new HashSet<>();
while (n != 1) {
int temp = 0;
while (n != 0) {
temp += Math.pow(n % 10, 2);
n /= 10;
}
if (set.contains(temp)) return false;
set.add(temp);
n = temp;
}
return true;
}
}
Python:
class Solution:
def isHappy(self, n: int) -> bool:
while 1:
if n == 4:
return False
if n == 1:
return True
n = sum([int(x) ** 2 for x in str(n)])
