lintcode 9 Fizz Buzz

题目描述

Description
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”.
  Example
  If n = 15, you should return:

  [
    “1”, “2”, “fizz”,
    “4”, “buzz”, “fizz”,
    “7”, “8”, “fizz”,
    “buzz”, “11”, “fizz”,
   “13”, “14”, “fizz buzz”
  ]

Challenge
Can you do it with only one if statement?
被3整除就fizz，被5整除就buzz，被3和5整除就fizz buzz
题目很简单，挑战很有意思，所以就记录一下。挑战让我们只用一个if来写

思路

这道题呢，当然如果用三元运算符一直嵌套也是可以的，不过明显不是这道题的意思
刚开始不知道怎么使用一个if来完美的解决这道题，后来看到一个老哥的解法，觉得还是比较有意思的
是这样的，写一个函数helper用来判断整除，然后用字符串拼接来输出字符串：

• 如果被3整除，字符串加上“fizz”
• 如果被15整除，加上“ ”
• 如果被5整除，加上“buzz”

String w = helper(i, 3, "fizz") + helper(i, 15, " ")
 + helper(i, 5, "buzz");

代码

public List<String> fizzBuzz(int n) {
        List<String> strings = new LinkedList<>();
        for (int i = 1; i <= n; i++) {
            String w = helper(i, 3, "fizz") + helper(i, 15, " ") + helper(i, 5, "buzz");
            if (w.equals("")) {
                w += i;
            }
            strings.add(w);
        }
        return strings;
    }

    public String helper(int n, int div, String w) {
        return n % div == 0 ? w : "";
    }