【leetcode】Happy Number（easy）

標籤：

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

思路：

這題目挺有意思的，肯定不能真的通過無限迴圈來判斷。我觀察了一下，估計不happy的數字，運算一圈後會出現前面已經出現過的數字，即在一組數字裡繞圈圈。故記錄一下出現過的數字，判斷是否重複。

如果重複了就不happy，得到1了就happy。

bool isHappy(int n) {        unordered_set<int> record;        while(1)        {            int sum = 0;            while(n > 0)            {                sum += (n % 10) * (n % 10);                n /= 10;            }            if(sum == 1)                return true;                        if(record.find(sum) == record.end()) //當前數字沒有出現過            {                record.insert(sum);                n = sum;            }            else                return false;        }    }

 

還有用O(1)空間的，就像鏈表找有沒有圈一樣，用快慢指標的思路。

public class Solution {    public boolean isHappy(int n) {        int x = n;        int y = n;        while(x>1){            x = cal(x) ;            if(x==1) return true ;            y = cal(cal(y));            if(y==1) return true ;            if(x==y) return false;        }        return true ;    }    public int cal(int n){        int x = n;        int s = 0;        while(x>0){            s = s+(x%10)*(x%10);            x = x/10;        }        return s ;    }}

 

還有用數學的，所有不happy的數字，都會得到4.（？？why）

bool isHappy(int n) {    if (n <= 0) return false;    int magic = 4;    while (1) {        if (n == 1) return true;        if (n == magic) return false;        int t = 0;        while (n) {            t += (n % 10) * (n % 10);            n /= 10;        }        n = t;    }}

 

【leetcode】Happy Number（easy）