Write a program, it outputs a number from 1 to n strings

Example:

n = 15,

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

Conventional thinking

Title very simple multiple output n is 3 Fizz, n is a multiple of output 5 Buzz, n while the output 3 or 5 multiples FizzBuzz. That is, when the determination condition, should first determine whether n is a multiple of 15, followed by determining whether a multiple of 3 or 5 to determine what output. code show as below:

class Solution {
    public List<String> fizzBuzz(int n) {
 List<String> list = new ArrayList<>(n);
 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(i + "");
 }
 }
 return list;
 }
}

Here to say, as if LeetCode use the List is not required when the lead pack in the code, I tried no matter did to the first line import java.util.*;no problem.

Below is the time and memory consumption:

Runtime: 1 ms
Memory Usage: 37.2 MB

Special ideas

Top is by %carried out modulo operation to achieve, I saw some people think the comments section:

Generally, for a CPU operand modulo relatively inefficient, it can be avoided if a large amount of remainder operands, can improve the performance of the program.

So there will be no use %of writing:

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 right; } }

Below is the time and memory consumption:

Runtime: 1 ms
Memory Usage: 37.3 MB

The program requires two variables in repeated self-energizing and re-assigned to 0, you can use the following programs to reduce these operations:

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

下面是时间与内存的消耗：

Runtime: 1 ms
Memory Usage: 37.1 MB

补充：`i+""`和`String.valueOf(i)`

which is better between
list.add( “” + i );
and
addStr = String.valueOf(i); list.add(addStr)？

我认为后者的写法更好，因为i+""实际上会new一个StringBuilder去拼接i和""，然后再调用toString()来得到字符串。而String.valueOf(i)在底层是调用了Integer.toString(i)来得到字符串。