30-Day Leetcoding Challenge Day2

本题涉及判断是否有loop

两种思路：1.用set存储路径 2.快慢指针法

JAVA：set()集合法

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> process = new HashSet<>();
        while(process.add(n)){
            int square = 0;
            while(n != 0){
                square += (n % 10)*(n % 10);
                n /= 10;
            }
            if(square == 1)
                return true;
            else
                n = square;
        }
        return false;
    }
}

JAVA：快慢指针法

class Solution {
    public boolean isHappy(int n) {
        int slow = n;
        int fast = square(n);
        while(fast != 1){
            slow = square(slow);
            fast = square(fast);
            fast = square(fast);
            if(slow == fast)
                return false;
        }
        return true;
    }
    public int square(int n){
        int res = 0;
        while(n != 0){
            res += (n%10)*(n%10);
            n /= 10;
        }
        return res;
    }
}

python3：set()集合法

class Solution:
    def isHappy(self, n: int) -> bool:
        process = set()
        while n != 1:
            n = sum([int(x)**2 for x in str(n)])
            if n in process:
                return False
            else:
                process.add(n)
        return True

python3：快慢指针法

class Solution:
    def isHappy(self, n: int) -> bool:
        slow = n
        fast = self.square(n)
        while fast != 1:
            slow = self.square(slow)
            fast = self.square(fast)
            fast = self.square(fast)
            if slow == fast:
                return False
        return True
    def square(self, n):
        res = 0
        while n:
            res += (n % 10)**2
            n //= 10
        return res