Code War

• code war honor 27
• uestc position 58

quesetion 1

Jaden Smith, the son of Will Smith, is the star of films such as The Karate Kid (2010) and After Earth (2013). Jaden is also known for some of his philosophy that he delivers via Twitter. When writing on Twitter, he is known for almost always capitalizing every word.

Your task is to convert strings to how they would be written by Jaden Smith. The strings are actual quotes from Jaden Smith, but they are not capitalized in the same way he originally typed them.

Example:

Not Jaden-Cased: "How can mirrors be real if our eyes aren't real"
Jaden-Cased: "How Can Mirrors Be Real If Our Eyes Aren't Real"
Note that the Java version expects a return value of null for an empty string or null.

translation 1

給一個初始的字符串，對于該字符串的每個單詞的首字母轉換為大寫并返回，當傳入字符串為空字符串(即"")或者傳入?yún)?shù)為null時，返回null。

public class JadenCase {

  public String toJadenCase(String phrase) {
    if(phrase == null || phrase.length() == 0 )
    {
        return null;
    }

    char[] str = phrase.toCharArray();
    str[0] = Character.toUpperCase(str[0]);
    for (int i = 0; i < str.length ; i++ )
    {
        if(str[i] == ' ')
        {
            str[i+1] = Character.toUpperCase(str[i+1]);
        }
    }
    return new String(str);
  }
}

1. 在判斷參數(shù)類型時，如果需要判斷傳入的參數(shù)是否為null， 則必須首先判斷是否為null，再判斷其他條件，反例： phrase.length() == 0 || phrase == null，在判斷這個條件時，如果phrase為null，這是首先判斷其長度，就會出錯。
2. 字符串轉化為字符串數(shù)組時，用str.CharArray();
3. 字符大小寫轉化：Character.toUpperCase(char c)、 Character.toLowerCase(char c)
4. 字符數(shù)組轉換為字符串 new String(char c[])

question 2

In a small town the population is p0 = 1000 at the beginning of a year. The population regularly increases by 2 percent per year and moreover 50 new inhabitants per year come to live in the town. How many years does the town need to see its population greater or equal to p = 1200 inhabitants?

At the end of the first year there will be:
1000 + 1000 * 0.02 + 50 => 1070 inhabitants

At the end of the 2nd year there will be:
1070 + 1070 * 0.02 + 50 => 1141 inhabitants (number of inhabitants is an integer)

At the end of the 3rd year there will be:
1141 + 1141 * 0.02 + 50 => 1213

It will need 3 entire years.

More generally given parameters:

p0, percent, aug (inhabitants coming or leaving each year), p (population to surpass)

the function nb_year should return n number of entire years needed to get a population greater or equal to p.

aug is an integer, percent a positive or null number, p0 and p are positive integers (> 0)

Examples:
nb_year(1500, 5, 100, 5000) -> 15
nb_year(1500000, 2.5, 10000, 2000000) -> 10
Note: Don't forget to convert the percent parameter as a percentage in the body of your function: if the parameter percent is 2 you have to convert it to 0.02.

translation 2

某個區(qū)域的人口初始為p0，每年以 百分之percent的速度增加，另外每年還會遷移過來aug個人,編程實現(xiàn)函數(shù)，返回幾年后人口會達到p。

class Arge {
    
    public static int nbYear(int p0, double percent, int aug, int p) {
        int sum;
        percent = percent*0.01;
        for(int i = 1; ; i++ )
        {
            sum = (int)(p0 + p0 * percent + aug);
            if( sum >= p)
            {
                return i;
            }
            p0 = sum;
        }
    }
}

1. 強制轉換 (int)float a;

question 3

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

For example:

 persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                      // and 4 has only one digit

 persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
                       // 1*2*6 = 12, and finally 1*2 = 2

 persistence(4) == 0 // because 4 is already a one-digit number

translation 3

實現(xiàn)一個函數(shù)persistence, 傳入一個正整數(shù)參數(shù)，把參數(shù)的每個數(shù)字相乘，得到的積如果不是個位數(shù)，那么重復上述操作，繼續(xù)乘，如果是個位數(shù)，那么返回計算的次數(shù)（即重復了多少次上述操作）

class Persist {
  public static int persistence(long n) {
    char str[] = (n+"").toCharArray();
        int time=0;
        while(str.length > 1)
        {
            int temp = 1;
            int[] key = new int[str.length];
            for(int i = 0; i < str.length; i++)
            {
                key[i] = Integer.parseInt(String.valueOf(str[i]));
                temp = key[i] * temp;
            }
            str = (temp+"").toCharArray();
            time++;
            if(temp < 10)
            {
                break;
            }

        }

        return time;
  }
}

1. 數(shù)字轉化為字符串 可使用如下方式 int + ""
2. Java中數(shù)組的創(chuàng)建方法: int[] arr = new int[length];
3. 字符轉化為數(shù)字：Integer.parseInt(String.valueOf(char c))
   即把'2'轉換為數(shù)字2，而強制轉換(int)char c;是把對應字符轉化為對應的ascii碼值

question 4

Take 2 strings s1 and s2 including only letters from ato z. Return a new sorted string, the longest possible, containing distinct letters,

each taken only once - coming from s1 or s2.

Examples:
a = "xyaabbbccccdefww" b = "xxxxyyyyabklmopq" longest(a, b) -> "abcdefklmopqwxy"
a = "abcdefghijklmnopqrstuvwxyz" longest(a, a) -> "abcdefghijklmnopqrstuvwxyz"

translation 4

給定兩個字符串，統(tǒng)計出現(xiàn)的字母并按照字母表的順序排序

public class TwoToOne {
    
    public static String longest (String s1, String s2) {
       char[] str = (s1+s2).toCharArray();
        int[] key = new int[26];
        char[] aim = new char[26];
        for(int i = 0; i < str.length; i++)
        {
            key[(int)str[i] - 97] = 1;
        }
        int i = 0;
        for(int j = 0; j < 26; j++)
        {
            if(key[j] == 1)
            {
                aim[i] = (char)(j+97);
                i++;
            }
        }
        char[] re = new char[i];
        System.arraycopy(aim, 0, re, 0,i);
        return new String(re);
    }
}

1. 字符串拼接 str1 + str2;
2. 數(shù)組復制函數(shù)
   arraycopy(@NotNull Object src, int srcPos,@NotNull Object dest,int destPos,int length)
   參數(shù) src 源數(shù)組，srcPos 源數(shù)組復制起點，dest 目標數(shù)組，destPos 目標數(shù)組開始位置，復制長度，

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