摘要：看了一些答案使用了多個(gè)這里給出幾種方案給大家參考下建立將所對(duì)應(yīng)的字符串分別映射，這樣一個(gè)即可對(duì)進(jìn)行優(yōu)化，只映射，這里需要一點(diǎn)數(shù)學(xué)技巧，大家看代碼就理解了利用的截?cái)喙δ軙簳r(shí)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(gè)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(duì)應(yīng)的字符串分別映射，這樣一個(gè)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(duì)1進(jìn)行優(yōu)化，只映射3,5,8，這里需要一點(diǎn)數(shù)學(xué)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?cái)喙δ?/p>
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(shí)想到這些，歡迎指正
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