leetcode 202 Happy Number
來源：程序員人生 發(fā)布時間：2015-09-23 08:18:17 閱讀次數(shù)：4143次
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1個代碼，用了些數(shù)論的知識吧：
https://leetcode.com/discuss/33014/4ms⑸-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 digits
def convert(self, n):
res = 0
while n > 0:
temp = n % 10
res += temp * temp
n = n // 10
return res
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