LeetCode Problem 66 : Plus One
$begingroup$
I have started practicing problems in leetcode and I solved the following problem on LeetCode.
The challenge is :
Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
I am trying to get better at writing good code for my solutions. Please give me your suggestions
class Solution {
public int plusOne(int a) {
int length = a.length-1;
while(length>=0)
{
a[length]=a[length]+1;
int carry = a[length]/10;
if(carry==1)
{
a[length]=0;
length=length-1;
}
else
break;
}
if(a[0]==0)
{
int array = new int[a.length+1];
array[0]=1;
for(int i=1;i<array.length-1;i++)
{
array[i]=0;
}
return array;
}
return a;
}
}
java programming-challenge
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
$endgroup$
add a comment |
$begingroup$
I have started practicing problems in leetcode and I solved the following problem on LeetCode.
The challenge is :
Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
I am trying to get better at writing good code for my solutions. Please give me your suggestions
class Solution {
public int plusOne(int a) {
int length = a.length-1;
while(length>=0)
{
a[length]=a[length]+1;
int carry = a[length]/10;
if(carry==1)
{
a[length]=0;
length=length-1;
}
else
break;
}
if(a[0]==0)
{
int array = new int[a.length+1];
array[0]=1;
for(int i=1;i<array.length-1;i++)
{
array[i]=0;
}
return array;
}
return a;
}
}
java programming-challenge
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
$endgroup$
add a comment |
$begingroup$
I have started practicing problems in leetcode and I solved the following problem on LeetCode.
The challenge is :
Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
I am trying to get better at writing good code for my solutions. Please give me your suggestions
class Solution {
public int plusOne(int a) {
int length = a.length-1;
while(length>=0)
{
a[length]=a[length]+1;
int carry = a[length]/10;
if(carry==1)
{
a[length]=0;
length=length-1;
}
else
break;
}
if(a[0]==0)
{
int array = new int[a.length+1];
array[0]=1;
for(int i=1;i<array.length-1;i++)
{
array[i]=0;
}
return array;
}
return a;
}
}
java programming-challenge
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
$endgroup$
I have started practicing problems in leetcode and I solved the following problem on LeetCode.
The challenge is :
Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
I am trying to get better at writing good code for my solutions. Please give me your suggestions
class Solution {
public int plusOne(int a) {
int length = a.length-1;
while(length>=0)
{
a[length]=a[length]+1;
int carry = a[length]/10;
if(carry==1)
{
a[length]=0;
length=length-1;
}
else
break;
}
if(a[0]==0)
{
int array = new int[a.length+1];
array[0]=1;
for(int i=1;i<array.length-1;i++)
{
array[i]=0;
}
return array;
}
return a;
}
}
java programming-challenge
java programming-challenge
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
edited 6 hours ago
Sᴀᴍ Onᴇᴌᴀ
9,55562164
9,55562164
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
asked 7 hours ago
Dhrubojyoti BhattacharjeeDhrubojyoti Bhattacharjee
14618
14618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The way you handle the extra digit for the last carry seems clumsy to me. To my mind, it would be much better to include the extra digit, by creating a new array at the start, then discard the extra digit if it's not needed.
Having that array for the answer, allows you to use it to hold the carry if needed.
When you know the numeric limits to your loop, it is much better to use a for loop rather than a while loop. This keeps the relative information for the loop in one place.
Putting this together it could look like this:
public int plusOne(int digits) {
int newLength = digits[0] == 9 ? digits.length + 1 : digits.length;
int answer = new int[newLength];
int a = newLength - 1;
answer[a] = 1;
for (int d = digits.length - 1; d >= 0; --d, --a) {
answer[a] += digits[d];
if (answer[a] == 10) {
answer[a] = 0;
answer[a-1] = 1;
}
}
return answer[0] > 0 ? answer : Arrays.copyOfRange(answer, 1, newLength);
}
answered 5 hours ago
tinstaafltinstaafl
6,7081928
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The way you handle the extra digit for the last carry seems clumsy to me. To my mind, it would be much better to include the extra digit, by creating a new array at the start, then discard the extra digit if it's not needed.
Having that array for the answer, allows you to use it to hold the carry if needed.
When you know the numeric limits to your loop, it is much better to use a for loop rather than a while loop. This keeps the relative information for the loop in one place.
Putting this together it could look like this:
public int plusOne(int digits) {
int newLength = digits[0] == 9 ? digits.length + 1 : digits.length;
int answer = new int[newLength];
int a = newLength - 1;
answer[a] = 1;
for (int d = digits.length - 1; d >= 0; --d, --a) {
answer[a] += digits[d];
if (answer[a] == 10) {
answer[a] = 0;
answer[a-1] = 1;
}
}
return answer[0] > 0 ? answer : Arrays.copyOfRange(answer, 1, newLength);
}
answered 5 hours ago
tinstaafltinstaafl
6,7081928
$endgroup$
add a comment |
$begingroup$
The way you handle the extra digit for the last carry seems clumsy to me. To my mind, it would be much better to include the extra digit, by creating a new array at the start, then discard the extra digit if it's not needed.
Having that array for the answer, allows you to use it to hold the carry if needed.
When you know the numeric limits to your loop, it is much better to use a for loop rather than a while loop. This keeps the relative information for the loop in one place.
Putting this together it could look like this:
public int plusOne(int digits) {
int newLength = digits[0] == 9 ? digits.length + 1 : digits.length;
int answer = new int[newLength];
int a = newLength - 1;
answer[a] = 1;
for (int d = digits.length - 1; d >= 0; --d, --a) {
answer[a] += digits[d];
if (answer[a] == 10) {
answer[a] = 0;
answer[a-1] = 1;
}
}
return answer[0] > 0 ? answer : Arrays.copyOfRange(answer, 1, newLength);
}
answered 5 hours ago
tinstaafltinstaafl
6,7081928
$endgroup$
add a comment |
$begingroup$
The way you handle the extra digit for the last carry seems clumsy to me. To my mind, it would be much better to include the extra digit, by creating a new array at the start, then discard the extra digit if it's not needed.
Having that array for the answer, allows you to use it to hold the carry if needed.
When you know the numeric limits to your loop, it is much better to use a for loop rather than a while loop. This keeps the relative information for the loop in one place.
Putting this together it could look like this:
public int plusOne(int digits) {
int newLength = digits[0] == 9 ? digits.length + 1 : digits.length;
int answer = new int[newLength];
int a = newLength - 1;
answer[a] = 1;
for (int d = digits.length - 1; d >= 0; --d, --a) {
answer[a] += digits[d];
if (answer[a] == 10) {
answer[a] = 0;
answer[a-1] = 1;
}
}
return answer[0] > 0 ? answer : Arrays.copyOfRange(answer, 1, newLength);
}
answered 5 hours ago
tinstaafltinstaafl
6,7081928
$endgroup$
The way you handle the extra digit for the last carry seems clumsy to me. To my mind, it would be much better to include the extra digit, by creating a new array at the start, then discard the extra digit if it's not needed.
Having that array for the answer, allows you to use it to hold the carry if needed.
When you know the numeric limits to your loop, it is much better to use a for loop rather than a while loop. This keeps the relative information for the loop in one place.
Putting this together it could look like this:
public int plusOne(int digits) {
int newLength = digits[0] == 9 ? digits.length + 1 : digits.length;
int answer = new int[newLength];
int a = newLength - 1;
answer[a] = 1;
for (int d = digits.length - 1; d >= 0; --d, --a) {
answer[a] += digits[d];
if (answer[a] == 10) {
answer[a] = 0;
answer[a-1] = 1;
}
}
return answer[0] > 0 ? answer : Arrays.copyOfRange(answer, 1, newLength);
}
answered 5 hours ago
tinstaafltinstaafl
6,7081928
answered 5 hours ago
tinstaafltinstaafl
6,7081928
answered 5 hours ago
tinstaafltinstaafl
6,7081928
answered 5 hours ago
tinstaafltinstaafl
6,7081928
6,7081928
add a comment |
add a comment |
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