[LeetCode] 412. Fizz Buzz 嘶嘶嗡嗡

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

But for 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"
]

很简单的一道题，最基本的思路就是对1~n的每一个数对3，5取模，根据情况写入结果。然后就是有一些极简的写法和思路，可以看看大牛们的写法，也挺受用的。

Java:  Not use '%' operation

public class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> ret = new ArrayList<String>(n);
        for(int i=1,fizz=0,buzz=0;i<=n ;i++){
            fizz++;
            buzz++;
            if(fizz==3 && buzz==5){
                ret.add("FizzBuzz");
                fizz=0;
                buzz=0;
            }else if(fizz==3){
                ret.add("Fizz");
                fizz=0;
            }else if(buzz==5){
                ret.add("Buzz");
                buzz=0;
            }else{
                ret.add(String.valueOf(i));
            }
        }
        return ret;
    }
}

Java:

public class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> list = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 && i % 5 == 0) {
                list.add("FizzBuzz");
            } else if (i % 3 == 0) {
                list.add("Fizz");
            } else if (i % 5 == 0) {
                list.add("Buzz");
            } else {
                list.add(String.valueOf(i));
            }
        }
        return list;
    }
}　　

Python:

def fizzBuzz(self, n):
    return ['Fizz' * (not i % 3) + 'Buzz' * (not i % 5) or str(i) for i in range(1, n+1)]

Python:　　

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

Python:

def fizzBuzz(self, n):
        return ['FizzBuzz'[i % -3 & -4:i % -5 & 8 ^ 12] or repr(i) for i in range(1, n + 1)]

Python:

class Solution(object):
    def fizzBuzz(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        result = []
        for i in xrange(1, n+1):
            if i % 15 == 0:
                result.append("FizzBuzz")
            elif i % 5 == 0:
                result.append("Buzz")
            elif i % 3 == 0:
                result.append("Fizz")
            else:
                result.append(str(i))

        return result

Python:　　

class Solution(object):
    def fizzBuzz(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        l = [str(x) for x in range(n + 1)]
        l3 = range(0, n + 1, 3)
        l5 = range(0, n + 1, 5)
        for i in l3:
            l[i] = 'Fizz'
        for i in l5:
            if l[i] == 'Fizz':
                l[i] += 'Buzz'
            else:
                l[i] = 'Buzz'
        return l[1:]　

Python: wo

class Solution(object):
    def fizzBuzz(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        res = []
        for i in xrange(1, n + 1):
            if i % 3 == 0 and i % 5 == 0:
                res.append('FizzBuzz')
            elif i % 3 == 0:
                res.append('Fizz')
            elif i % 5 == 0:
                res.append('Buzz')
            else:
                res.append(str(i))

        return res

C++:

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> res;
        for (int i = 1; i <= n; ++i) {
            if (i % 15 == 0) res.push_back("FizzBuzz");
            else if (i % 3 == 0) res.push_back("Fizz");
            else if (i % 5 == 0) res.push_back("Buzz");
            else res.push_back(to_string(i));
        }
        return res;
    }
};

　　

　　

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