备注:此篇答案中有部分没有获取到3个Skill Rating。回头有时间再改进。
02.01
public class Program {
public static int[] Puzzle(int n) {
int[] array = new int[n];
for(int i=0 ; i < n; i++){
array[i] = i;
}
return array;
}
}
02.02
public class Program {
public static int[] Puzzle(int n) {
int[] array = new int[n];
for(int i=0 ; i < n; i++){
array[i] = i * n;
}
return array;
}
}
02.03
public class Program {
public static int[] Puzzle(int n) {
int[] array = new int[n];
for(int i=0 ; i < n; i++){
array[i] = i * i;
}
return array;
}
}
02.04
public class Program {
public static int Puzzle(int[] v) {
int total = 0;
for(int i :v){
total += i;
}
return total;
}
}
02.05
public class Program {
public static int Puzzle(int n) {
int total = 0;
for(int i = 0 ; i < n ; i++){
total += i*i;
}
return total;
}
}
02.06
public class Program {
public static int Puzzle(String s) {
int j = 0;
for(int i = 0 ; i < s.length() ; i ++){
if (s.charAt(i).equals('a'))
j++;
}
return j;
}
}
02.07
public class Program {
public static int Puzzle(String s, char x) {
int j = 0;
for(int i = 0 ; i < s.length() ; i ++){
if (s.charAt(i).equals(x))
j++;
}
return j;
}
}
PS:02.06 和02.07 已优化,可得三星。说实话,在玩Code Hunt 之前,从来没想过查找字符串中的字符次数竟然可以这样,给Code Hunt 一个赞。
//02.06
public class Program {
public static int Puzzle(String s) {
String s2 = s.replace("a","");
return s.length()-s2.length();
}
}
//02.07
public class Program {
public static int Puzzle(String s, char x) {
String s2 = s.replace(x+"","");
return s.length()-s2.length();
}
}
PS:再度优化02.05
public class Program {
public static int Puzzle(String s, char x) {
return (n*(n-1)*(2*n-1))/6;
}
}