412. Fizz Buzz Fizz Buzz

https://leetcode.com/problems/fizz-buzz/open in new window

• Total Accepted: 31093
• Total Submissions: 53272
• Difficulty: Easy

Question

Write a program that outputs the string representation of numbers from 1 to n.

Butfor multiples of three it should output “Fizz” instead of the number and for the multiples of five output “Buzz”. For numbers which are multiples of both three and five output “FizzBuzz”.

Example:

n = 15,

Return:
[
    "1",
    "2",
    "Fizz",
    "4",
    "Buzz",
    "Fizz",
    "7",
    "8",
    "Fizz",
    "Buzz",
    "11",
    "Fizz",
    "13",
    "14",
    "FizzBuzz"
]

题目大意

从1~n这么多数字中中，如果某个位置是3的倍数，把这个数字换成Fizz，如果是5的倍数，把这个数字换成Buzz，如果既是3的倍数又是5的倍数，换成FizzBuzz.

解题方法

方法一：遍历判断

思路很简单，判断是否能特定位置的数字是否能被3和5整除即可。

class Solution(object):
    def fizzBuzz(self, n):
    	"""
    	:type n: int
    	:rtype: List[str]
    	"""
    	ListReturn = [];
    	x = 1
    	while x <= n:
    		if x % 3 == 0 and x % 5 == 0:
    			ListReturn.append("FizzBuzz")
    		elif x % 3 == 0:
    			ListReturn.append("Fizz")
    		elif x % 5 == 0:
    			ListReturn.append("Buzz")
    		else:
    			ListReturn.append(str(x))
    		x += 1
    	return ListReturn

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AC:69 ms

感觉好繁琐，python应该可以很简单。所以参考了别人的跟进如下。

class Solution(object):
    def fizzBuzz(self, n):
    	"""
    	:type n: int
    	:rtype: List[str]
    	"""
    	return ["Fizz" * (i % 3 == 0) + "Buzz" * (i % 5 == 0) 
    	+ str(i) * (i % 3 != 0 and i % 5 != 0)			
    	for i in range(1, n + 1)]

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AC:96 ms

嗯。这个看起来舒服多了。

方法二：字符串相加

如果是5的倍数，就把结果字符串后面加上Buzz即可。这里不能使用elif的判断，因为是15既是3的倍数又是5的倍数，所以需要加上两个字符串。

class Solution:
    def fizzBuzz(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        res = []
        for i in range(1, n + 1):
            pos = ""
            if i % 3 == 0:
                pos += "Fizz"
            if i % 5 == 0:
                pos += "Buzz"
            if not pos:
                pos = str(i)
            res.append(pos)
        return res

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方法三：字典

把方法二的判断进行了优化，使用字典保存3和5的字符串的结果对应。

class Solution:
    def fizzBuzz(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        res = []
        strmap = {3 : "Fizz", 5 : "Buzz"}
        for i in range(1, n + 1):
            pos = ""
            for j in [3, 5]:
                if i % j == 0:
                    pos += strmap[j]
            if not pos:
                pos = str(i)
            res.append(pos)
        return res

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