[LeetCode] 202. Happy Number 快乐数

时间:2023-02-12 02:20:50

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

Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.

写一个算法判断一个数是不是快乐数。快乐数:一个正整数,如果对其各个位上的数字分别平方求和得到一个新的数,再进行同样的操作,如果结果变成了1,则说明是快乐数。

解法: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,那以后会一直循环,不可能达到1。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。

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;
}
}

Python:

class Solution:
# @param {integer} n
# @return {boolean}
def isHappy(self, n):
lookup = {}
while n != 1 and n not in lookup:
lookup[n] = True
n = self.nextNumber(n)
return n == 1 def nextNumber(self, n):
new = 0
for char in str(n):
new += int(char)**2
return new  

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

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;
}
};

  

All LeetCode Questions List 题目汇总