leetcode 202 Happy Number

標籤：c++   algorithm   leetcode   

 

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 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 = 1

我的解決方案：

// happy number.cpp : Defines the entry point for the console application.//#include "stdafx.h"#include <iostream>#include <set>using namespace std;  bool isHappy(int n)    {        int split = 0;        int sum = 0;                set<int> myset;        set<int>::iterator it;                while(sum != 1)        {            do            {                split = n % 10;                n = n / 10;                sum = split* split + sum;            }while(n>0);                            if(sum==1)            {                break;            }            else            {                                    it=myset.find(sum);if(it!=myset.end())                {                    return false;                }                myset.insert(sum);n = sum;sum = 0;            }        }                return true;    }int _tmain(int argc, _TCHAR* argv[]){isHappy(19);return 0;}

最短的一個代碼，用了些數論的知識吧：

https://leetcode.com/discuss/33014/4ms-5-line-c-code

bool isHappy(int n) {    while(n>6){        int next = 0;        while(n){next+=(n%10)*(n%10); n/=10;}        n = next;    }    return n==1;}

兩個python代碼：



 def isHappy(self, n):    return self.isHappyHelper(n, {})def isHappyHelper(self, n, prev):    if n == 1:        return True    elif n not in prev:        prev[n] = 1    else:        return False    new = 0    for char in str(n):        new += int(char)**2    return self.isHappyHelper(new, prev)

class Solution:# @param {integer} n# @return {boolean}def isHappy(self, n):    table = []    n = self.convert(n)    while n != 1:        if n in table:            return False        else:            table.append(n)            n = self.convert(n)    return True# @param {integer} n# @return {integer} sum of digitsdef convert(self, n):    res = 0    while n > 0:        temp = n % 10        res += temp * temp        n = n // 10    return res



leetcode 202 Happy Number