I just need lines 6 through 9. Here are the lines I currently have: Line 1: if (s.length() == 0) { Line 2: return 0; Line 3: if (s.charAt(0) == ‘1’) { Line 4: return 1 + countOnes(s.substring(1)); Line 5: return countOnes(s.substring(1));
I just need lines 6 through 9. Here are the lines I currently have: Line 1: if (s.length() == 0) { Line 2: return 0; Line 3: if (s.charAt(0) == ‘1’) { Line 4: return 1 + countOnes(s.substring(1)); Line 5: return countOnes(s.substring(1));
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
Related questions
Question
I just need lines 6 through 9.
Here are the lines I currently have:
Line 1: if (s.length() == 0) {
Line 2: return 0;
Line 3: if (s.charAt(0) == ‘1’) {
Line 4: return 1 + countOnes(s.substring(1));
Line 5: return countOnes(s.substring(1));
![LINE1
V [ Choose ]
if (s.length() == 0) {
if (s.charAt(0) == '0') {
LINE2
return countOnes(s.substring(0));
return '1';
if (s.length == 1) {
LINE3
if (s.length() :
1) {
} else {
if (s.charAt[0] == '0') {
LINE4
}
return "";
return 0;
LINE5
return countOnes(s.substring(1);
return 1+ countOnes(s.substring(1));
if (s.charAt(0) == '1') {
LINE6
{
return countOnes(s.substring(2));
if (s.length
== 0) {
LINE7
return 1+ countOnes(s.substring(2));
return '0';
return 1;
LINE8
return 1+ countOnes(s.substring(0));
if (s.charAt[0] == '1') {](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd97c0ef-a3a9-4c4a-b75b-2f4b31d21cac%2Ff16d6ea8-f00b-433a-b55b-caba9a21500b%2F21xn72q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:LINE1
V [ Choose ]
if (s.length() == 0) {
if (s.charAt(0) == '0') {
LINE2
return countOnes(s.substring(0));
return '1';
if (s.length == 1) {
LINE3
if (s.length() :
1) {
} else {
if (s.charAt[0] == '0') {
LINE4
}
return "";
return 0;
LINE5
return countOnes(s.substring(1);
return 1+ countOnes(s.substring(1));
if (s.charAt(0) == '1') {
LINE6
{
return countOnes(s.substring(2));
if (s.length
== 0) {
LINE7
return 1+ countOnes(s.substring(2));
return '0';
return 1;
LINE8
return 1+ countOnes(s.substring(0));
if (s.charAt[0] == '1') {
![Counting the number of 1 bits in a bit string s can be accomplished in Java by first initializing an
integer counter n to 0. Next, we implement a loop which iterates over each character c of s and
when c is '1' we add 1 to n. When c is '0' we do nothing. Here is the implementation of that
algorithm:
private int countOnes(String s) {
int n = 0;
for (int i = 0; i < s.length(); ++i) {
if (s.charAt(i) == '1') {
++n;
}
return n;
This problem is also amenable to being solved using a recursive method rather than an iterative
method (one that employs a loop). Your job is to select the correct pieces of code from those
available and arrange them in proper order to implement a recursive method with the same method
signature.
private int countOnes(String s) {
[LINE1]
[LINE2]
[LINE3]
[LINE4]
[LINE5]
[LINE6]
[LINE7]
[LINE8]
[LINE9]
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd97c0ef-a3a9-4c4a-b75b-2f4b31d21cac%2Ff16d6ea8-f00b-433a-b55b-caba9a21500b%2Fr4b90tn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Counting the number of 1 bits in a bit string s can be accomplished in Java by first initializing an
integer counter n to 0. Next, we implement a loop which iterates over each character c of s and
when c is '1' we add 1 to n. When c is '0' we do nothing. Here is the implementation of that
algorithm:
private int countOnes(String s) {
int n = 0;
for (int i = 0; i < s.length(); ++i) {
if (s.charAt(i) == '1') {
++n;
}
return n;
This problem is also amenable to being solved using a recursive method rather than an iterative
method (one that employs a loop). Your job is to select the correct pieces of code from those
available and arrange them in proper order to implement a recursive method with the same method
signature.
private int countOnes(String s) {
[LINE1]
[LINE2]
[LINE3]
[LINE4]
[LINE5]
[LINE6]
[LINE7]
[LINE8]
[LINE9]
}
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