摘要：看了一些答案使用了多个这里给出几种方案给大家参考下建立将所对应的字符串分别映射，这样一个即可对进行优化，只映射，这里需要一点数学技巧，大家看代码就理解了利用的截断功能暂时想到这些，欢迎指正
9.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".
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?
看了一些答案使用了多个if,这里给出几种方案给大家参考下
1、建立HashMap,将3,6,9,12,5,10,0(i%15==0)所对应的字符串分别映射，这样一个if(map.containsKey(i%15))即可
public static HashMap map = new HashMap<>();
static{
map.put(3,"fizz");
map.put(6,"fizz");
map.put(9,"fizz");
map.put(12,"fizz");
map.put(5,"buzz");
map.put(10,"buzz");
map.put(0,"fizz buzz");
}
public List fizzBuzz(int n) {
List ret = new ArrayList<>();
for(int i=1;i<=n;i++){
int key = i%15;
if(map.containsKey(key)){
ret.add(map.get(key));
}else{
ret.add(""+i);
}
}
return ret;
}
2、对1进行优化，只映射3,5,8，这里需要一点数学技巧，大家看代码就理解了
public static HashMap map = new HashMap<>();
static{
map.put(3,"fizz");
map.put(5,"buzz");
map.put(8,"fizz buzz");
}
public List fizzBuzz(int n) {
List ret = new ArrayList<>();
for(int i=1;i<=n;i++){
int key = f(i,3)+f(i,5);
if(map.containsKey(key)){
ret.add(map.get(key));
}else{
ret.add(""+i);
}
}
return ret;
}
private int f(int i,int k){
return (k-i%k)*((k-i%k)/k);
}
3、利用&&的截断功能
public List fizzBuzz(int n) {
List ret = new ArrayList<>(n);
for(int i=1;i<=n;i++){
String str = null;
boolean bool1 = i%3==0&&(str="fizz")!=null;
boolean bool2 = i%5==0&&(str="buzz")!=null&& i%3==0&&(str="fizz buzz")!=null;
if(str==null){
ret.add(""+i);
}else {
ret.add(str);
}
}
return ret;
}
暂时想到这些，欢迎指正
文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。
转载请注明本文地址：https://www.ucloud.cn/yun/77101.html
