Coursera scala课程第一周答案

Exercise 1: Pascal’s Triangle

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

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

    def pascal(c: Int, r: Int): Int = {
      if(c==r||c==0)
        1
      else
        pascal(c - 1, r - 1) + pascal(c, r - 1)
  }

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.

    def balance(chars: List[Char]): Boolean = {
      def balanced(chars: List[Char], open: Int): Boolean = {
        if (chars.isEmpty) open == 0
        else if (chars.head == '(') balanced(chars.tail,open+1)
        else if (chars.head == ')') open>0 && balanced(chars.tail,open-1)
        else balanced(chars.tail,open)
      }
      balanced(chars,0)
    }

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.

Hint: Think of the degenerate cases. How many ways can you give change for 0 CHF(swiss money)? How many ways can you give change for >0 CHF, if you have no coins?

    def countChange(money: Int, coins: List[Int]): Int = {
      if(money == 0)
        1
      else if(money > 0 && !coins.isEmpty)
        countChange(money - coins.head, coins) + countChange(money, coins.tail)
      else
        0
    }