leetcode筆記：Happy Number

標籤：leetcode   c++   map   algorithm   演算法   

一. 題目描述

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 （各個位的平方和）
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

二. 題目分析

題目要求對任意一個正整數，不斷各個數位上數位平方和，若經過若干次運算後結果收斂到1，則該數字為Happy Number，不是Happy Number的數在經過多次運算後會從某個數開始陷入迴圈。這道題目我們只用根據規則進行計算，並使用一個map儲存已經出現過的數字，這樣每輪計算完成後尋找map，若發現值已存在，證明已陷入迴圈，可跳出迴圈並判定該整數不是Happy Number。

更多關於Happy Number的知識可參考：http://baike.baidu.com/link?url=slZGeshN-Igmd4geJ7PZ9hPwk_PZGZK6QttvzH5TpUiQIPy8qZAmG6o9-5-x-Eu2WGoQuhnASB7alb2ecEatpVj0C9-3DaQPjy0Cpvmvp7AB3IiKy6vu1Qb8FhCbj_Eg1Nt4ba9FDzZT1BzbQcsL5a

三. 範例程式碼

#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) // 陷入迴圈，判為非快樂數                return false;            else            {                showNum[temp] = true;                n = temp;            }        }    }};

一些測試結果：

四. 小結

無

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leetcode筆記：Happy Number