摘要：看了一些答案使用了多個這里給出幾種方案給大家參考下建立將所對應的字符串分別映射，這樣一個即可對進行優化，只映射，這里需要一點數學技巧，大家看代碼就理解了利用的截斷功能暫時想到這些，歡迎指正
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個if,這里給出幾種方案給大家參考下
1、建立HashMap,將3,6,9,12,5,10,0(i%15==0)所對應的字符串分別映射，這樣一個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、對1進行優化，只映射3,5,8，這里需要一點數學技巧，大家看代碼就理解了
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&&的截斷功能
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;
}
暫時想到這些，歡迎指正
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