[leetcode]: 202. Happy Number

1.题目

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
判断一个数是否为happy number.
happy number的定义为对这个数的所有位上的数求平方和得到新的数，重复这个过程，如果能得到数字1，说明为happy number。

2.分析

自己随手写几个数就会发现一个规律，

例如18
1+64=65
36+25=61
36+1=37
9+49=58
25+64=89
64+81=145
1+16+25=42
16+4=20
4+0=4
16
1+36=37
…

如果一个数不是happy number，那么上述求解过程会陷入一个死循环。即形成了一个cycle。

判断是够会形成cycle，有两种方式：
1）建一个hashTable，存储出现过的数字，每次判断得到的新数是否已经出现过。
2）floyd cycle detection algorithm（solution里看到的）
两个指针slow每次走1步,fast每次走2(or more)步。如果存在cycle，那么slow和fast必定在某点相遇。

3.代码

floyd方法

class Solution {public:    int getNext(int n) {        int sum = 0;        while (n > 0) {            sum += (n % 10)*(n % 10);            n /= 10;        }        return sum;    }    bool isHappy(int n) {        int slow = n;        int fast = n;        do {            slow = getNext(slow);            fast = getNext(getNext(fast));        } while (slow != fast);        return slow == 1;    }};

hashTable方法

bool isHappy(int n) {    set<int> appear;    appear.insert(n);    while (true) {        int tmp = 0;        while (n > 0) {            tmp += (n % 10)*(n % 10);            n /= 10;        }        n = tmp;        cout << tmp << " ";        if(tmp==1)            return true;        if (appear.find(tmp) != appear.end())//有数字重复出现，说明是死循环            break;        else            appear.insert(tmp);    }    return false;}