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"]

分析

遍历1~n的所有整数并以字符串形式存入List中，3的倍数"Fizz"，5的倍数"Buzz"，15的倍数"FizzBuzz"

解答

解法1：（我）（3ms√）

public class Solution {    public List
    
      fizzBuzz(int n) {        List
     
       list = new ArrayList
      
       ();        for (int i = 1; i <= n; i++) {            if (i % 15 == 0) {                list.add("FizzBuzz");            }            else if (i % 3 == 0) {                list.add("Fizz");            }            else if (i % 5 == 0) {                list.add("Buzz");            }            else{                list.add(Integer.toString(i));//或String.valueOf(i)            }        }        return list;    }}
      
     
    

 

解法2：不用"%"操作符（4ms）

Integer.highestOneBit() 返回一个int值：如果i具有'1'位，则返回值具有1个'1'位，其位置即是i的最高位（最左边）的'1'位的位置；如果i不具有'1'位，则i=0，返回0。例：Integer.highestOneBit(5) = 4，因为5的二进制表示为101，返回值的二进制表示为100

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