Exercise 1: Pascal’s Triangle

The following pattern of numbers is called Pascal’s triangle.

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1
   ...

The numbers at the edge of the triangle are all 1, and each number inside the triangle is the sum of the two numbers above it. Write a function that computes the elements of Pascal’s triangle by means of a recursive process.

Do this exercise by implementing the pascal function in Main.scala, which takes a column c and a row r, counting from 0 and returns the number at that spot in the triangle. For example, pascal(0,2)=1,pascal(1,2)=2 and pascal(1,3)=3.

def pascal(c: Int, r: Int): Int

Exercise 2: Parentheses Balancing

Write a recursive function which verifies the balancing of parentheses in a string, which we represent as a List[Char] not a String. For example, the function should return true for the following strings:

(if (zero? x) max (/ 1 x))
I told him (that it’s not (yet) done). (But he wasn’t listening)

The function should return false for the following strings:

:-)
())(

The last example shows that it’s not enough to verify that a string contains the same number of opening and closing parentheses.

Do this exercise by implementing the balance function in Main.scala. Its signature is as follows:

def balance(chars: List[Char]): Boolean

There are three methods on List[Char] that are useful for this exercise:

• chars.isEmpty: Boolean returns whether a list is empty
• chars.head: Char returns the first element of the list
• chars.tail: List[Char] returns the list without the first element
  Hint: you can define an inner function if you need to pass extra parameters to your function.

Testing: You can use the toList method to convert from a String to aList[Char]: e.g. "(just an) example".toList.

Exercise 3: Counting Change

Write a recursive function that counts how many different ways you can make change for an amount, given a list of coin denominations. For example, there are 3 ways to give change for 4 if you have coins with denomination 1 and 2: 1+1+1+1, 1+1+2, 2+2.

Do this exercise by implementing the countChange function inMain.scala. This function takes an amount to change, and a list of unique denominations for the coins. Its signature is as follows:

def countChange(money: Int, coins: List[Int]): Int

Once again, you can make use of functions isEmpty, head and tail on the list of integers coins.

Code

/**
 * Exercise 1
 */
def pascal(c: Int, r: Int): Int = {
    //  满足条件始终返回1
    if (c == 0 || r == c)
        1
    else
        pascal(c - 1, r - 1) + pascal(c, r - 1)
}

/**
 * Exercise 2
 */
def balance(chars: List[Char]): Boolean = {
    @scala.annotation.tailrec
    def loop(chars: List[Char], cnt: Int): Boolean = {
        if (cnt < 0)
            false
        else if (chars.isEmpty && cnt == 0)
            true
        else {
            if (chars.head == '(')
                loop(chars.tail, cnt + 1)
            else if (chars.head == ')')
                loop(chars.tail, cnt - 1)
            else
                loop(chars.tail, cnt)
        }
    }

    loop(chars, 0)
}

/**
 * Exercise 3
 */
def countChange(money: Int, coins: List[Int]): Int = {
    if (money < 0 || coins.isEmpty)
        0
    else if (money == 0)
        1
    else    // 很好的形式，值得借鉴
        countChange(money - coins.head, coins) + countChange(money, coins.tail)
}