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;
    }
};