Lintcode9-Fizz Buzz-Easy

Fizz Buzz

Given number n. Print number from 1 to n. But:

• when number is divided by 3, print "fizz".
• when number is divided by 5, print "buzz".
• when number is divided by both 3 and 5, print "fizz buzz".
• when number can‘t be divided by either 3 or 5, print the number itself.

Example

If n = 15, you should return:
[
  "1", "2", "fizz",
  "4", "buzz", "fizz",
  "7", "8", "fizz",
  "buzz", "11", "fizz",
  "13", "14", "fizz buzz"
]

If n = 10, you should return:
[
  "1", "2", "fizz",
  "4", "buzz", "fizz",
  "7", "8", "fizz",
  "buzz"
]

Challenge

Can you do it with only one if statement?

注意：

1. 同时被3和5整除，应该是 num % 15 == 0 , 而不是 nums % 3 == 0 && nums % 5 == 0 , 后者会加长运行时间。
2. 四种情况的判断顺序很重要，同样影响运行时间。采用 "if - else if - else if - else":  if （num % 15 == 0）- else if (num % 5 == 0) - else if (num % 3 == 0) - else. else if 对判断先后顺序有要求，不可颠倒。而如果全部是if, 每个if判断的条件会增加，即增加运行时间。if （num % 15 == 0）- if (num % 5 == 0 && num % 3 != 0) -if (num % 3 == 0 && num % 5 != 0) - else.
3. Java int -> String : String s = String.valueOf(i);
4. List.add()  参数只能是字符串，所以 line 13.

代码：

 1 public List<String> fizzBuzz(int n) {
 2         int num = 1;
 3         List<String> fblist = new ArrayList<String>();
 4         
 5         while (num <= n){
 6             if (num % 15 == 0) {
 7                 fblist.add("fizz buzz");
 8             } else if (num % 5 == 0) {
 9                 fblist.add("buzz");
10             } else if (num % 3 == 0) {
11                 fblist.add("fizz");
12             } else {
13                 fblist.add(String.valueOf(num));
14             }
15             num++;
16         }
17         return fblist;
18     }