leetcode 202. 快乐数(Happy Number)

编写一个算法来判断一个数是不是“快乐数”。

一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和，然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那么这个数就是快乐数。

示例: 

输入: 19
输出: true
解释: 
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

  原题链接

在没想到更好到方法到时候暴力也许是最好的方法，如果1000次循环都不变为1，很可能他就是死循环了

另外不要局限于一种实现，可以适当的把下面贴的代码都看一看

class Solution {
    public boolean isHappy(int n) {
    	int time=1000;
    	while(time-->0){
    		String s = ""+n;
    		int ans = 0;
    		for(int i=0;i<s.length();i++)
    			ans+=Integer.parseInt(""+s.charAt(i))*Integer.parseInt(""+s.charAt(i));
    		if(ans==1)
    			return true;
    		n=ans;
    	}
        return false;
    }
}

做一个set的小改动，不需要到1000次循环才判断为死循

Beat 90% Fast Easy Understand Java Solution with Brief Explanation 

The runtime now beats 76% but still thanks for the brief discussion and easy to understand code.

public boolean isHappy(int n) {
    Set<Integer> inLoop = new HashSet<Integer>();
    int squareSum,remain;
	while (inLoop.add(n)) {
		squareSum = 0;
		while (n > 0) {
		    remain = n%10;
			squareSum += remain*remain;
			n /= 10;
		}
		if (squareSum == 1)
			return true;
		else
			n = squareSum;

	}
	return false;

}

beat 100% java in China

class Solution {
    public boolean isHappy(int num) {
        if(num < 10) {
            return num == 1 || num == 7;
        }
        int n = num;
        int b = 0;
        while(n > 0) {
            b += (n % 10) * (n % 10);
            n = n / 10;
        }
        return isHappy(b);
    }
}

  beat 95+% java

class Solution {
    public boolean isHappy(int n) {
        if(n == 1) return true;
        boolean res = false;
        HashSet<Integer> hashSet = new HashSet<>();
        hashSet.add(n);
        while(true){
            int sum = 0;
            while(n > 0){
                int temp = n % 10;
                n /= 10;
                sum += temp * temp;
            }
            if(sum == 1){
                res = true;
                break;
            }
            n = sum;
            if(hashSet.contains(n)) return false;
            hashSet.add(n);
        }
        return res;
    }
}

神奇数学法，java版

class Solution {
    public boolean isHappy(int n) {
        int slow = n, fast = n;
        while (fast != 1) {
            fast = digitSquareSum(fast);
            fast = digitSquareSum(fast);
            slow = digitSquareSum(slow);
            if (slow == fast) break;
        }
        return fast == 1;
    }
    
    private int digitSquareSum(int n) {
        int res = 0;
        while (n != 0) {
            res += (n % 10) * (n % 10);
            n /= 10;
        }
        return res;
    }
}

神奇数学法，C版

I see the majority of those posts use hashset to record values. Actually, we can simply adapt the Floyd Cycle detection algorithm. I believe that many people have seen this in the Linked List Cycle detection problem. The following is my code:

我看到大多数帖子都使用hashset来记录值。实际上，我们可以简单地调整Floyd Cycle检测算法。我相信许多人已经在Linked List Cycle检测问题中看到了这一点。以下是我的代码：

int digitSquareSum(int n) {
    int sum = 0, tmp;
    while (n) {
        tmp = n % 10;
        sum += tmp * tmp;
        n /= 10;
    }
    return sum;
}

bool isHappy(int n) {
    int slow, fast;
    slow = fast = n;
    do {
        slow = digitSquareSum(slow);
        fast = digitSquareSum(fast);
        fast = digitSquareSum(fast);
    } while(slow != fast);
    if (slow == 1) return 1;
    else return 0;
}