[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

Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.

public class Solution {
    public boolean isHappy(int n) {
        List<Integer> list = new ArrayList<Integer>();

        if (n == 0)
            return false;
        if (n == 1)
            return true;

        int temp = 0, sum = 0;

        while (true) {
            temp = n % 10;
            sum = sum + temp * temp;
            n = n / 10;
            if (n == 0) {
                if (sum == 1)
                    return true;
                if (list.contains(sum))
                    return false;
                n = sum;
                list.add(sum);
                sum = 0;
            }
        }
    }
}

给两个别人的写法。

1.空间换时间

public boolean isHappy(int n) {  
        if(n==0){  
            return false;  
        }  

        if(n==1){  
            return true;  
        }  

        int temp=0;  
        boolean flag[]=new boolean[811];  
        int yushu=0;  

        while(true){  
            while(n!=0){  
                yushu=n%10;  
                n=n/10;  
                temp+=yushu*yushu;  
            }  

            if(temp==1){  
                return true;  
            }  

            if(flag[temp]){  
                return false;  
            }else{  
                 flag[temp]=true;  
            }  

            n=temp;  
            temp=0;  
        }  
    }  

2.大神之作。。。位运算，反正我是不会

public boolean isHappy(int n) {  
    int[] mark = new int[8];  
    while (n > 1) {  
        n = convert(n);  
        if (n < 243) {  
            int sec = n >> 5;  
            int mask = 1 << (n & 0x1f);  
            if ((mark[sec] & mask) > 0) {  
                return false;  
            }  
            mark[sec] |= mask;  
        }  
    }  
    return true;  
}  

private int convert(int n) {  
    int sum = 0;  
    while (n > 0) {  
        int t = n % 10;  
        sum += t * t;  
        n /= 10;  
    }  
    return sum;  
}