leetcode 202 Happy Number
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
我的解决方案:
// 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; }
最短的一个代码,用了些数论的知识吧:
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; }
两个python代码:
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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