leetcode 202. Happy Number

Posted 2020-10-30 将者，智、信、仁、勇、严也。

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

很自然的解法是：

class Solution(object):
    def isHappy(self, n):
        """
        :type n: int
        :rtype: bool
        """        
        def sum2(num):
            ans = 0
            while num != 0:
                ans += (num % 10)**2
                num = num/10
            return ans
        
        m = set()
        while n not in m:
            if n == 1:
                return True
            m.add(n)
            n = sum2(n)
        return False

 还有可以使用环路链表的判断方法，一个快慢指针一起赛跑，直到追上：

class Solution(object):
    def isHappy(self, n):
        """
        :type n: int
        :rtype: bool
        """        
        def sum2(num):
            ans = 0
            while num != 0:
                ans += (num % 10)**2
                num = num/10
            return ans
        
        slow = n
        fast = sum2(sum2(n))
        while slow != fast:
            slow = sum2(slow)
            fast = sum2(sum2(fast))                
        return slow == 1

 java代码其实非常优雅：

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;
}

 数学解法：

class Solution {
public:
int loop[8] = {4,16,37,58,89,145,42,20};

bool inLoop(int n){
    for(auto x: loop){
        if(x == n) return true;
    }
    return false;
}

bool isHappy(int n) {
    if(n == 1) return true;
    if(inLoop(n)) return false;
    int next = 0;
    while(n){
        next += (n%10)*(n%10);
        n /= 10;
    }
    return isHappy(next);
}

};

 

proof:

1.loop number is less than 162.

Assume f(x) is the sum of the squares of x’s digits. let’s say 0 < x <= 9,999,999,999 which is larger than the range of an int. f(x) <= 9^2 * 10 = 810. So no mater how big x is, after one step of f(x), it will be less than 810.The most large f(x) value (x < 810) is f(799) = 211. And do this several times: f(211) < f(199) = 163. f(163) < f(99) = 162. So no mater which x you choose after several times of f(x),it finally fall in the range of [1,162] and can never get out.

2.I check every unhappy number in range of [1,162] , they all fall in loop {4,16,37,58,89,145,42,20} ,which means every unhappy number fall in this loop.

其实通过枚举就应该知道上述<=810范围内的数字都会落在特定范围，通过不断缩小范围来找规律。=》

Using fact all numbers in [2, 6] are not happy (and all not happy numbers end on a cycle that hits this interval):

bool isHappy(int n) {
    while(n>6){
        int next = 0;
        while(n){next+=(n%10)*(n%10); n/=10;}
        n = next;
    }
    return n==1;
}

class Solution(object):
    def isHappy(self, n):
        """
        :type n: int
        :rtype: bool
        """        
        def sum2(num):
            ans = 0
            while num != 0:
                ans += (num % 10)**2
                num = num/10
            return ans
        
        while n > 6:
            n = sum2(n)
        return n == 1