简单数组训练

任务一

返回数组 array [1, 2, 3, 4, 5, 6]

/** Returns an array [1, 2, 3, 4, 5, 6] */  
public static int[] makeDice() {  
    // TODO: Fill in this function.  
    return new int[]{1, 2, 3, 4, 5, 6};  
}

任务二

* Returns the order depending on the customer.
* If the customer is Ergun, return ["beyti", "pizza", "hamburger", "tea"]. * If the customer is Erik, return ["sushi", "pasta", "avocado", "coffee"]. * In any other case, return an empty String[] of size 3.

public static String[] takeOrder(String customer) {  
   // TODO: Fill in this function.  
   if ("Ergun".equals(customer)) {  
       return new String[]{"beyti", "pizza", "hamburger", "tea"};  
   } else if ("Erik".equals(customer)) {  
       return new String[]{"sushi", "pasta", "avocado", "coffee"};  
   } else {  
       return new String[3];  
   }  
}

任务三

返回给定数组中最大元素和最小元素之间的正差值。假定给定的数组是非空的

public static int findMinMax(int[] array) {  
    // TODO: Fill in this function.  
    int min = array[0];  
    int max = array[0];  
    for (int i : array) {  
        if (i > max) {  
            max = i;  
        }  
        if (i < min) {  
            min = i;  
        }  
    }  
    return max - min;
}

思路是新建两个变量分别储存最大值与最小值，遍历后更新变量的值

任务四 冰雹序列

目标: 使用递归来计算从输入数字 n 开始的冰雹序列 (hailstone sequence)，并将该序列以一个整数列表（List<Integer>）的形式返回。

冰雹序列的规则如下：

1. 从一个正整数 n 开始。

2. 如果 n 是偶数，则将 n 除以 2。

3. 如果 n 是奇数，则将 n 乘以 3 再加 1。

4. 重复这个过程，直到 n 变为 1。

private static List<Integer> hailstoneHelper(int x, List<Integer> list) {  
    // TODO: Fill in this function.  
    list.add(x);  
    if (x == 1){  
        return list;  
    }  
  
    if (x % 2 ==0) {  
        return  hailstoneHelper(x / 2, list);  
    } else {  
        return hailstoneHelper(x * 3 + 1, list);  
    }  
}

采用了递归的思想，根据正课中的 IntList 即可理解链表、递归的思想

ListExpercises

任务一：返回总和

/** Returns the total sum in a list of integers */

public static int sum(List<Integer> L) {  
    // TODO: Fill in this function.  
    int sum = 0;  
    for (int num : L){  
        sum += num;  
    } return sum;  
}

任务二：返回列表内偶数

/** Returns a list containing the even numbers of the given list */

public static List<Integer> evens(List<Integer> L) {  
    // TODO: Fill in this function.  
  List<Integer> evens = new ArrayList<>();  
  for (int num : L){  
      if (num % 2 == 0){  
        evens.add(num);  
      }  
  } return evens;  
}

任务三： 返回两列表内相同的数

/** Returns a list containing the common item of the two given lists */ 第一种，嵌套循环

public static List<Integer> common(List<Integer> L1, List<Integer> L2) {  
    List<Integer> common = new ArrayList<>();  
    for (int num1 : L1){  
        for (int num2 : L2){  
            if (num1 == num2){  
                common.add(num1);  
                break;  
            }  
        }  
    } return common;  
}

第二种，contains

public static List<Integer> common(List<Integer> L1, List<Integer> L2){
	List<Integer> common = new Arrary<>();
	for (int num1 : L1){
		if (L2.contains(num1)){
			common.add(num1)
		}
	} return common;

}

任务四 返回字符 c 在列表出现的次数

public static int countOccurrencesOfC(List<String> words, char c) {  
    // TODO: Fill in this function.  
    int times = 0;  
    for (String word : words) {  
        for (char letter : word.toCharArray()) {  
            if (letter == c) {  
            times++;  
            }  
        }  
    } return times;  
}

还可以使用charAt进行查找

MapExercise

任务一 字母数字映射表

Returns a map from every lower case letter to the number corresponding to that letter, where 'a' is
1, 'b' is 2, 'c' is 3, ..., 'z' is 26.

public static Map<Character, Integer> letterToNum() {  
    // TODO: Fill in this function.  
    Map<Character, Integer> letterToNum = new HashMap<>();  
    int num = 1;  
    for (char letter = 'a'; letter <= 'z'; letter++) {  
        letterToNum.put(letter, num);  
        num++;  
    }  
      
    return letterToNum;  
}

采用了 for 循环对 char 进行连续赋值

任务二 返回平方

Returns a map from the integers in the list to their squares. For example, if the input list
is [1, 3, 6, 7], the returned map goes from 1 to 1, 3 to 9, 6 to 36, and 7 to 49.

public static Map<Integer, Integer> squares(List<Integer> nums) {  
    // TODO: Fill in this function.  
    Map<Integer, Integer> squares = new HashMap<>();  
    for (Integer num1 : nums) {  
        squares.put(num1, num1 * num1);  
    }  
  
    return squares;  
}

任务三 返回字母出现次数

Returns a map of the counts of all words that appear in a list of words.

public static Map<String, Integer> countWords(List<String> words) {  
    // TODO: Fill in this function.  
    Map<String, Integer> wordCount = new HashMap<>();  
    for (String word : words) {  
        if (wordCount.containsKey(word)) {  
            wordCount.put(word, wordCount.get(word) + 1);  
        } else {  
            wordCount.put(word, 1);  
        }  
    }  
  
    return wordCount;  
}