[英]Can someone help me understand these 2 pieces of code [Recursion]?
我很難理解這段代碼中發生的事情。 在針對作業的14個問題中,這是我唯一遇到的兩個問題,只是在不知道答案如何的情況下給出答案。 任何幫助,將不勝感激,謝謝。
這是#1:
public void printDollarSign(int k)
{
int j;
if (k>0)
{
for (j=1; j<= k; j++)
System.out.print("$");
System.out.println();
printDollarSign(k-1);
}
}
如果調用是,輸出將是什么:printDollarSign(5); ?
$$$$$
$$$$
$$$
$$
$
這是#2:
public void bbb(String s, int p)
{
if (p>= 0)
{
bbb(s,p-1);
System.out.print(s.charAt(p));
}
}
如果調用是,輸出將是什么:bbb(“ January”,4); ?
亞努阿
printDollarSign的說明:
對於首次通話:
printDollarSign(k)首先打印k個$,然后打印換行,然后調用printDollarSign(k-1)首先打印k-1個$,然后打印換行,然后調用printDollarSign(k -1-1)...一直持續到k = 0。 當k = 0時,printDollarSign(0)不打印任何內容。
您的第一個方法是創建一個帶有int特定參數的函數,在本例中為“ k”。
碼:
public void printDollarSign(int k){ //Define function
int j;//Define j as an int
if (k>0){//Check if k is greater than 0
for (j=1; j<= k; j++)//Loop through k where j = 1
System.out.print("$");//Print $ by the amount of k
System.out.println();//Print a new line
printDollarSign(k-1);//Re run the function
}
}
您的第二個問題是創建一個帶有字符串和int“ s”和“ p”兩個參數的函數
碼:
public void bbb(String s, int p){//Define Function
if (p>= 0){ //Check if p is greater than 0
bbb(s,p-1);//Rerun function
System.out.print(s.charAt(p));//Print character of string based on p
}
}
添加代碼注釋以進行解釋:
1號
public void printDollarSign(int k)
{
// temporary variable j, it will be always initialized to 1 inside the for loop.
int j;
// only executed to be true if k is more than 0, that means if K is initially 5
// then it only works for 5,4,3,2,1 and not for 0.
if (k>0)
{
// always starts with 1 and goes to the value of k, that means if K is currently 5
// then it will print 5 dollars, if 4 then 4 dollars and so on
for (j=1; j<= k; j++)
System.out.print("$");
// a new line will be added once dollars are printed.
System.out.println();
// this will again call the same function and decrements the value of k, so next time
// k will have the value one less then the previous one, if this has printed 5 dollars
// in last iteration next time it will print 4 dollars and so on
printDollarSign(k-1);
}
}
2號
public void bbb(String s, int p)
{
// only print a character if p has a value grater than 0. in you case , p has a value 4 that
// mean only 4 characters will be printed at max
if (p>= 0)
{
// recuresively call same method by decrementing p, so it will be
// [bbb(s,3) prints 'a']-> [bbb(s,3) prints 'u']-> bbb(s,2) [prints 'n']-> [bbb(s, 1) prints 'a']> [bbb(s, 0) prints 'J']
// last character will be printed first
bbb(s,p-1);
// prints the character at p location
System.out.print(s.charAt(p));
}
}
bbb的解釋:
執行:
現在彈出每個元素並在該位置打印字符
if you provide a string s as 'January', the way it stores is like this:
0th position = J,
1th position = a,
2th position = n,
3th position = u,
4th position = a,
5th position = r,
6th position = y,
Just like an array of characters.
When you provide p=4, you are setting the closing criteria for the condition check.
function bbb('January', 4){
if(4>=0){
bbb('January', 3); ....
function bbb('January', 3){
if(3>=0){
bbb('January', 2); ....
function bbb('January', 2){
if(2>=0){
bbb('January', 1); ....
function bbb('January', 1){
if(1>=0){
bbb('January', 0); ....
function bbb('January', 0){
if(0>=0){
bbb('January', -1); ....
function bbb('January', -1){
if(-1>=0){ Here condition check fails.. hence char at(-1) doec not print
return goes back to print of previous call and prints J as p=0, we have J
the p=1: a
p=2: n
p=3: u
p=4: a
功能完成...
請讓我知道說明是否有幫助
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