leetcode-202 快乐数 Python

最新推荐文章于 2021-03-15 23:14:44 发布

Xenonon

最新推荐文章于 2021-03-15 23:14:44 发布

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CC 4.0 BY-SA版权

文章标签： leetcode 动态规划算法 python 数据结构

本文链接：https://blog.youkuaiyun.com/Xenonon/article/details/105868552

本文介绍了三种解决LeetCode快乐数问题的方法，包括使用集合、双指针和数学技巧，详细阐述了每种方法的思路和代码实现。

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https://leetcode-cn.com/problems/happy-number/solution/kuai-le-shu-by-leetcode-solution/

方法一：集合

class Solution:
    def isHappy(self, n: int) -> bool:
        if n < 0:
            return False
        seen = set()
        while n not in seen:
            seen.add(n)
            nex = 0
            for digit in str(n):
                nex += int(digit) ** 2
            if nex == 1:
                return True
            n = nex
        return False

def isHappy(self, n: int) -> bool:

    def get_next(n):
        total_sum = 0
        while n > 0:
            n, digit = divmod(n, 10)
            total_sum += digit ** 2
        return total_sum

    seen = set()
    while n != 1 and n not in seen:
        seen.add(n)
        n = get_next(n)

    return n == 1

方法二：双指针

class Solution:
    def isHappy(self, n: int) -> bool:
        def get_next(number):
            total_sum = 0
            while number > 0:
                number, digit = divmod(number, 10)
                total_sum += digit ** 2
            return total_sum
    
        slow = n
        fast = get_next(n)
        while fast != 1 and slow != fast:
            slow = get_next(slow)
            fast = get_next(get_next(fast))
        return fast == 1

方法三：数学

class Solution:
    def isHappy(self, n: int) -> bool:
        nums = {4, 16, 37, 58, 89, 145, 42, 20}
        def get_next(number):
            total_sum = 0
            while number > 0:
                number, digit = divmod(number, 10)
                total_sum += digit ** 2
            return total_sum
        while n != 1 and n not in nums:
            n = get_next(n)
        return n == 1