leetcode笔记:Happy Number
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一. 题目描述
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 (各个位的平方和)
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
二. 题目分析
题目要求对任意一个正整数,不断各个数位上数字的平方和,若经过若干次运算后结果收敛到1,则该数字为Happy Number,不是Happy Number的数在经过多次运算后会从某个数开始陷入循环。这道题目我们只用根据规则进行计算,并使用一个map存储已经出现过的数字,这样每轮计算完成后查找map,若发现值已存在,证明已陷入循环,可跳出循环并判定该整数不是Happy Number。
更多关于Happy Number的知识可参考:http://baike.baidu.com/link?url=slZGeshN-Igmd4geJ7PZ9hPwk_PZGZK6QttvzH5TpUiQIPy8qZAmG6o9-5-x-Eu2WGoQuhnASB7alb2ecEatpVj0C9-3DaQPjy0Cpvmvp7AB3IiKy6vu1Qb8FhCbj_Eg1Nt4ba9FDzZT1BzbQcsL5a
三. 示例代码
#include <iostream>#include <map>using namespace std;class Solution{public: bool isHappy(int n) { if (n < 0) return false; if (n == 1) return true; map<int, bool> showNum; while (true) { int temp = 0; while (n) { temp += (n % 10) * (n % 10); n /= 10; } if (temp == 1) return true; else if (showNum[temp] == true) // 陷入循环,判为非快乐数 return false; else { showNum[temp] = true; n = temp; } } }};
一些测试结果:
四. 小结
无
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