大概是上周开始玩CodeWars的，反正就是觉得蛮有意思的，编程刷荣誉。一开始等级都是8kyu，经过一个星期的摸爬滚打，本鸟升到了7kyu哈哈，记录下这个过程吧。（注：Java版本1.8.0_91 (Java 8)）
1、Square Every Digit
Description:
Welcome. In this kata, you are asked to square every digit of a number.
For example, if we run 9119 through the function, 811181 will come out.
Note: The function accepts an integer and returns an integer
我的答案：
public class SquareDigit {
public int squareDigits(int n) {
String str = String.valueOf(n);
int length = str.length();
String s=new String();
for (int i = 0; i < length; i++) {
char b = str.charAt(i);
int num = Integer.valueOf(b + "");
num = num * num;
s=s+num;
}
int numder=Integer.valueOf(s);
return numder;
}
}
获赞最多的答案：
public class SquareDigit {
public int squareDigits(int n) {
String result = "";
while (n != 0) {
int digit = n % 10 ;
result = digit*digit + result ;
n /= 10 ;
}
return Integer.parseInt(result) ;
}
}
哪个答案更优雅这显而易见，我这就不再自我吐槽了。
2、Compare Strings by Sum of Chars
Description:
Compare two strings by comparing the sum of their values (ASCII character code).
For comparing treat all letters as UpperCase.
Null-Strings should be treated as if they are empty strings.
If the string contains other characters than letters, treat the whole string as it would be empty.
Examples:
"AD","BC" -> equal
"AD","DD" -> not equal
"gf","FG" -> equal
"zz1","" -> equal
"ZzZz", "ffPFF" -> equal
"kl", "lz" -> not equal
null, "" -> equal
Your method should return true, if the strings are equal and false if they are not equal.
我的答案：
public class Kata
{
public static Boolean compare(String s1, String s2)
{
int strLength01 = 0;
int strLength02 = 0;
if (s1!=null) {
strLength01=s1.length();
}
if (s2!=null) {
strLength02=s2.length();
}
int sum01 = 0;
int sum02 = 0;
boolean hasOther01 = false;
boolean hasOther02 = false;
for (int i = 0; i < strLength01; i++) {
int asii = s1.charAt(i);
if ((asii >= 'a' && asii <= 'z') || (asii >= 'A' && asii <= 'Z')) {
if (asii >= 'a') {
asii -= 32;
}
sum01 += asii;
} else// 有其他字符
{
hasOther01=true;
System.out.println("s1 has other char");
if (s2==null||s2.equals("")) {
System.out.println("s2 是null或者empet");
return true;
}
}
}
System.out.println();
for (int i = 0; i < strLength02; i++) {
int asii = s2.charAt(i);
if ((asii >= 'a' && asii <= 'z') || (asii >= 'A' && asii <= 'Z')) {
if (asii >= 'a') {
asii -= 32;
}
sum02 += asii;
}
else// 有其他字符
{
hasOther02=true;
if (s1==null||s1.equals("")) {
return true;
}
}
}
System.out.println();
if(hasOther01&&hasOther02)
{
return true;
}
if ((sum01 == sum02)&&!hasOther01&&!hasOther02) {
return true;
}
return false;
}
}
推荐答案：
public class Kata {
public static boolean compare(String s1, String s2) {
if (s1 == null || !s1.matches("[a-zA-Z]+")) s1 = "";
if (s2 == null || !s2.matches("[a-zA-Z]+")) s2 = "";
return s1.toUpperCase().chars().sum() == s2.toUpperCase().chars().sum();
}
}
我咋就没想到用正则呢？？？不说了，让本鸟哭会儿......
3、Mumbling
Description:
This time no story, no theory. The examples below show you how to write function accum:
Examples:
Accumul.accum("abcd"); // "A-Bb-Ccc-Dddd"
Accumul.accum("RqaEzty"); // "R-Qq-Aaa-Eeee-Zzzzz-Tttttt-Yyyyyyy"
Accumul.accum("cwAt"); // "C-Ww-Aaa-Tttt"
The parameter of accum is a string which includes only letters from a..z and A..Z.
我的答案：
public class Accumul {
public static String accum(String s) {
// your code
int strLength = s.length();
String s1 = new String();
StringBuffer s3=new StringBuffer();
for (int i = 0; i < strLength; i++) {
StringBuffer s2 = new StringBuffer();
char ch = s.charAt(i);
for (int j = 0; j <= i; j++) {
if (j == 0) {
ch = Character.toUpperCase(ch);
s2.append(ch);
}
else
{
ch = Character.toLowerCase(ch);
s2.append(ch);
}
}
s1=s2.toString();
if(i==strLength-1)
{
s3.append(s1);
}
else
s3.append(s1+"-");
}
return s3.toString();
}
}
推荐答案：
public class Accumul {
public static String accum(String s) {
StringBuilder bldr = new StringBuilder();
int i = 0;
for(char c : s.toCharArray()) {
if(i > 0) bldr.append('-');
bldr.append(Character.toUpperCase(c));
for(int j = 0; j < i; j++) bldr.append(Character.toLowerCase(c));
i++;
}
return bldr.toString();
}
}
4、Descending Order
Description:
Your task is to make a function that can take any non-negative integer as a argument and return it with it's digits in descending order. Essentially, rearrange the digits to create the highest possible number.
Examples:
Input: 21445 Output: 54421
Input: 145263 Output: 654321
Input: 1254859723 Output: 9875543221
我的答案：
public class DescendingOrder {
public static int sortDesc(final int num) {
//Your code
String s=num+"";
int length=s.toString().length();
int[] array=new int[length];
int nu=1;
int rest=0;
for (int i = 1; i <= array.length; i++) {
for (int j = array.length-i; j >=1; j--) {
nu=nu*10;
}
if(i==1)
{
System.out.println(nu);
array[i-1]=num/nu;
rest=num%nu;
System.out.println("rest "+rest);
}
else
{
System.out.println(nu);
array[i-1]=rest/nu;
rest=rest%nu;
System.out.println("rest "+rest);
}
nu=1;
}
int key;
int sa=0;
//开始排序
for (int i = 1; i < array.length; i++) {
key=array[i];
sa=i-1;
while (sa>=0&&array[sa]<key) {
array[sa+1]=array[sa];
sa--;
}
array[sa + 1] = key;
}
String number="";
for (int i = 0; i < array.length; i++) {
number=number+array[i];
}
int numb=Integer.valueOf(number);
System.out.println(numb+"");
return numb;
}
}
推荐答案：
import java.util.Comparator;
import java.util.stream.Collectors;
public class DescendingOrder {
public static int sortDesc(final int num) {
return Integer.parseInt(String.valueOf(num)
.chars()
.mapToObj(i -> String.valueOf(Character.getNumericValue(i)))
.sorted(Comparator.reverseOrder())
.collect(Collectors.joining()));
}
}
5、Sum of odd numbers
Description:
Given the triangle of consecutive odd numbers:
1
3 5
7 9 11
13 15 17 19
21 23 25 27 29
...
Calculate the row sums of this triangle from the row index (starting at index 1) e.g.:
rowSumOddNumbers(1); // 1
rowSumOddNumbers(2); // 3 + 5 = 8
推荐答案：
class RowSumOddNumbers {
public static int rowSumOddNumbers(int n) {
return n * n * n;
}
}
6、Multiples of 3 and 5
Description:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in.
Note: If the number is a multiple of both 3 and 5, only count it once.
Courtesy of ProjectEuler.net
推荐答案：
public class Solution {
public int solution(int number) {
int sum=0;
for (int i=0; i < number; i++){
if (i%3==0 || i%5==0){sum+=i;}
}
return sum;
}
}
