LeetCode 202. Happy Number(快乐数)
题目描述
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
解法一
思路
“求一个数的所有数字的平方和”挺简单的:
sum = 0
while n:
sum += (n % 10)**2
n //= 10
按照题目的描述,需要循环求平方和,即求出当前数组的平方和后,再以此平方和作为新的数继续求平方和,而循环终止条件是:得到的平方和为1
,或得到的平方和在之前的循环中出现过。判断平方和是否为1
很简单,每次检查就好了;而判断平方和是否出现过,则只需要维持一个Set,每次循环检查当前平方和是否在Set中,在则终止循环,不在则将此平方和放到Set中。
Python
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
got = set()
while n != 1 and n not in got:
got.add(n)
sum = 0
while n:
sum += (n % 10)**2
n //= 10
n = sum
return n == 1
Java
public class Solution {
public boolean isHappy(int n) {
Set<Integer> got = new HashSet<>();
while (n != 1 && !got.contains(n)) {
got.add(n);
int sum = 0;
while (n != 0) {
sum += Math.pow(n % 10, 2);
n /= 10;
}
n = sum;
}
return n == 1;
}
}
C++
class Solution {
public:
bool isHappy(int n) {
set<int> got;
while (n != 1 && got.find(n) == got.end()) {
got.insert(n);
int sum = 0;
while (n) {
sum += pow(n % 10, 2);
n /= 10;
}
n = sum;
}
return n == 1;
}
};