20170306-leetcode-202-Happy Number

2019-11-06 06:24:10

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1.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: **19 is a happy number

1² + 9² = 828² + 2² = 686² + 8² = 1001² + 0² + 0² = 1 https://leetcode.com/problems/happy-number/?tab=Description

解读给定任何一个正整数，计算每位数上的平方，求和，并不断重复计算，如果有1出现，则为happyNumber，否则会进入一个循环；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。 http://baike.baidu.com/link?url=Xc-p5vE0DPUwPkbEj7qeOjTlVSY3TLq-fCUEs-t3gBLQ9DjEWpajg55YCtTP539DABsp6coRd_GCzggVRwA79hsm8dxy4cbIYo3tiGyw8KLXAP6PX-XRio8BfmdbeWc8NZkJKA51b0FMhL7KiYZOjtdA1BTBcbFgc4JZSDcSIxK

2.Solution

class Solution(object): def isHappy(self, n): """ :type n: int :rtype: bool """ stack = [] n=self.caculate(n) stack.append(n) while 1: n=self.caculate(n) if n==1:return True else: if n in stack:return False else: stack.append(n) def caculate(self,n): return sum([int(x)**2 for x in str(n)])

把计算的记过存放在一个stack中，如果中间结果又有1的直接返回true，如果中间结果等于stack中的某一个值，那么返回false

class Solution(object): def isHappy(self, n): n=self.caculate(n) while 1: n=self.caculate(n) if n==1:return True if n==4:return False def caculate(self,n): return sum([int(x)**2 for x in str(n)])

如果计算的过程中有4的出现，那么返回false

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