LeetCode – Reverse Words in a String II (Java)

Category: Algorithms May 12, 2014

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.

The input string does not contain leading or trailing spaces and the words are always separated by a single space.

For example,
Given s = "the sky is blue",
return "blue is sky the".

Could you do it in-place without allocating extra space?

Java Solution

public void reverseWords(char[] s) {
    int i=0;
    for(int j=0; j<s.length; j++){
        if(s[j]==' '){
            reverse(s, i, j-1);        
            i=j+1;
        }
    }
 
    reverse(s, i, s.length-1);
 
    reverse(s, 0, s.length-1);
}
 
public void reverse(char[] s, int i, int j){
    while(i<j){
        char temp = s[i];
        s[i]=s[j];
        s[j]=temp;
        i++;
        j--;
    }
}

Category >> Algorithms

If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:

<pre><code> 
String foo = "bar";
</code></pre>

sly

The solution can definitely be simplified with nested loop and 4 pointers. e.g. i and j for nested loop, start and end for each word to reverse, i for string end-to-end, j for words separator. Swap chars between start and end until floor(stat/end) for each word. Reset start and end according I and J to swap next word.
Neha Kundra

def reverse_string(string) : # split the string string = string.split(' ') n = len(string) - 1 rev_string = '' while n >= 0 : rev_string = rev_string + (string[n]) + ' ' n = n - 1 return rev_string
if __name__ == '__main__' : string = input("Enter the string: ") print(reverse_string(string).strip())
Neha Kundra

https://uploads.disquscdn.com/images/cf02e6653b054797a297d4a7911b8e1539315433418f3ff821244b40c44b1e71.png
Luis Gerardo Camara Salinas

I did like this (split is a token function):
std::string reverseWords(std::string str){ std::vector list = split(str," "); std::string r_string; std::reverse(list.rbegin(), list.rend()); for(auto x:list){ r_string+=x+" "; }; return r_string; }
Sahar Goelman

in python this is a one liner
print(' '.join([word[::-1] for word in input.split(' ')]))
saipabbathi

Your first method is taking up the stack space though in recursion.
Mohamed

public static String reverseWord(String word) {

String [] splitting = word.split(” “);
String [] tempStringArray = new String[splitting.length];

for(int i=0;i<tempStringArray.length;i++)
{
tempStringArray[i] = splitting[splitting.length-i-1];
}

return String.join(" ",tempStringArray);
}
Godfred Sabbih

String reverseString(String [ ]s, int index){ if(index>=s.lenght) return "";
return s[0]+" "+reverseString(s,index+1); }
Michael Brooks

Python3:

def reverseString(str):
temp = str.split(” “)
res = “”
for i in range(len(temp)):
res += temp[~i] + ” ”
return res[:-1]
Evelyn Barr

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My id 457887
Devendra Vaja

Inline without extra space
String reverseWords2(String input) { int i = input.lastIndexOf(" "); if(i == -1) { //no space left, this is the first word return input; }
return input.substring(i+1) + " " + reverseWords2(input.substring(0, i)); }

WIth extra space but readable

String reverseWords(String input) { if(input.isEmpty()) { return input; }
//split input into words String[] wa = input.split(" "); StringBuilder sb = new StringBuilder(); //create a new string buffer and fill it from backwards for(int i = wa.length-1; i >=0; i--) { sb.append(wa[i]); sb.append(" "); } return sb.toString().trim(); }
Devendra Vaja

Inline solution no extra space with recursion
String reverseWords2(String input) { System.out.println(input); int i = input.lastIndexOf(" "); if(i == -1) { //no space left, this is the first word return input; }
return input.substring(i+1) + " " + reverseWords2(input.substring(0, i)); }

Solution with extra space

String reverseWords(String input) { //Read input //split it into words if(input.isEmpty()) { return input; }
String[] wa = input.split(" "); StringBuilder sb = new StringBuilder(); //create a new string buffer and fill it from backwards for(int i = wa.length-1; i >=0; i--) { sb.append(wa[i]); sb.append(" "); } return sb.toString().trim();
}
James Alikhani

with python3:

def reverse(A):
n = len(A)
B = []
for i in range(n):
B.append(A.pop())
return B

def slice_reverse(A, j, k):
A[j:k] = reverse(A[j:k])

def reverseWords(S):
S = list(S)
n = len(S)
j=0
for i in range(n):
if S[i]==’ ‘:
slice_reverse(S, j, i)
j=i+1
slice_reverse(S, j, n+1)
return ”.join(reverse(S))

S = “the sky is blue”
reverseWords(S)
rhel.user

Java 8 solution:

public static void rev1(String[] a) { String[] result = Stream.iterate(a.length - 1, i -> i - 1) .limit(a.length) .map(i -> a[i]) .toArray(String[] :: new);
Arrays.stream(result).forEach(System.out::println); }
Perminder

One line code in JavaScript:
function reverseString(str){ return str.split(' ').reverse().join(' '); } var str = "the sky is blue"; var output = reverseString(str); document.write(output);
Boris Afanasiev

Minus one line:
public static void reverseWords(char[] s) { int i=0; for(int j=0; j<=s.length; j++){ if(j == s.length || s[j]==' '){ reverse(s, i, j-1); i=j+1; } }
// reverse(s, i, s.length-1);
reverse(s, 0, s.length-1); }
Brandon

using System.Linq;
var output = string.Join(” “, s.Split(‘ ‘).Reverse());

#dotnetrocks 😉
Ayman Ali

var input = “ayman ali”;
input = ” ” + input;
var arr = input.ToCharArray();
var output = string.Empty;
var b = 0;
var e = arr.Length;
for (int p = arr.Length – 1; p >= 0; p–)
{
if (arr[p] == ‘ ‘ )// little off beginning of word
{
b = p + 1;
output += input.Substring(b, e – b) + ” “;
e = p ;

}
}
Console.WriteLine(output.TrimEnd(‘ ‘));
Console.ReadLine();
vijay

package java8fileread;

public class Rev {

public static void main(String[] args) {
String s = “hello”;
String rev = reverse(s.toCharArray(), ‘ ‘, s.length());
System.out.println(rev);
}

static String rev= “”;
private static String reverse(char[] charArray, char e, int length) {
if(length != 0)
{
if(e != ‘ ‘) rev += e;
reverse(charArray, charArray[–length], length) ;
}
return rev + charArray[0];
}
}
tan nguyen

public static String reverse(String s) {
String tmp = new StringBuilder(s).reverse().toString();
char[] in = tmp.toCharArray();
int j = 0;
for (int i = 0; i < in.length; i++) {
if (in[i] == ' ') {
reverseChar(in, j, i – 1);
j = i + 1;
}
if (i == in.length-1)
{
reverseChar(in, j, i);
}
}
StringBuilder sb = new StringBuilder(in.length);
sb.append(in);

return sb.toString();
}

public static void reverseChar(char[] input, int i, int j) {
int b = j;
for (int a = i; a <= b; a++) {
char tmp = input[a];
input[a] = input[b];
input[b] = tmp;
b–;
}
//final StringBuilder sb = new StringBuilder(input.length);
//sb.append(input);

// System.out.print(sb.toString());
}
Sagarika Dusane

String word = “The sky is Blue”;
List wordList = Arrays.asList(word.split(” “));
Collections.reverse(wordList);
word = String.join(” “, (CharSequence[]) wordList.toArray());

System.out.println(word);
}
Eldar

Two variant of solution using C#

class Program
{
static string reverseString1(string input)
{
char temp;
char[] s = input.ToCharArray();
for (int i = 0; i < s.Length / 2; temp = s[s.Length – 1 – i] , s[s.Length – 1 – i] = s[i] , s[i++] = temp ) ;
return new string(s);
}

static string reverseString2(string input)
{
char[] s = input.ToCharArray();
for (int i = 0; i < s.Length / 2; s[s.Length – 1 – i] ^= s[i], s[i] ^= s[s.Length – 1 – i], s[s.Length – 1 – i] ^= s[i++]) ;
return new string(s);
}

static void Main(string[] args)
{
string input = "abcd";
Console.WriteLine(reverseString1(input));
Console.WriteLine(reverseString2(input));
}
}
Bukary Kandeh

Trimming will get rid of the space between the words, that’s NOT what we want to do. I append space between the words so “sb.length() – 1” just deletes the last space.

Example: String s = “the sky is blue”;

*If you apply the TRIM you’ll get the following: “blueisskythe”.
*My application WITHOUT the “sb.length() – 1”: “blue is sky the ”
*My application WITH the “sb.length() – 1”: “blue is sky the”

setLength() is the most efficient way to get rid of the final space. Basically, it alters the count and overwrites the value we don’t want with a zero byte.
Carlos Hernandez

how about trimming the SB instead of setting the length?
Carlos Pérez

funny, I had done something really similar:
public static String reverseWords(String str) {
StringBuilder sb = new StringBuilder();
int limit = str.length();
//from last to first
for (int i = str.length() – 1; i > 0; i–) {
if (str.charAt(i) == ‘ ‘) {
sb.append(str.substring(i + 1, limit)).append(‘ ‘);
limit = i;
}
}
sb.append(str.substring(0, limit));

return sb.toString();
}
Brikesh Kumar

public class Program { public static void Main(string[] args) { //Your code goes here var input = "Hello"; Console.WriteLine(string.Format("Before reversing: - {0}",input)); var inPutArray = input.ToCharArray(); //var reversed = new StringBuilder(); //Reverse the string for(int i= 0, j = inPutArray.Length - 1; i <= j;i++,j--){
var temp = inPutArray[i]; inPutArray[i] = inPutArray[j]; inPutArray[j] = temp; } Console.WriteLine(string.Format("After reversing: - {0}",new String(inPutArray))); } }
Bukary Kandeh

public static void ReverseString(String s) { String[] split = s.split(" "); StringBuilder sb = new StringBuilder();
for (int i = split.length - 1; i >= 0; i--) { sb.append(split[i] + " "); } System.out.println(sb.setLength(sb.length() - 1).toString()); }
karthik339

I liked this implementaion. Fast and efficinent.
Huy

I think from the moment you have array = string.split(” “); you are already using O(n) space, where n is the number of ‘ ‘ in the string.
Dnyanada Pramod Arjunwadkar

String s="the sky is blue"; String[] s1=s.split("\s");
List l1=new ArrayList(); l1=Arrays.asList(s1); Collections.reverse(l1); System.out.println("this-:"+l1); StringBuilder sb= new StringBuilder(); for(int y=0;y<l1.size();y++) { if(y==l1.size()-1) { sb.append(l1.get(y)); } else { sb.append(l1.get(y)+" "); } } String hy=sb.toString(); hy.trim(); System.out.print(hy);
sewa

i think using a stringbuilder or stringbuffer would be better as they expand and contract as opposed to Strings that are immutable. this way, you dont keep allocating new spaces for every new append you do.
bpgen

String s = “the sky is blue”;
String[] splitted = s.split(” “);
System.out.println(Arrays.stream(splitted).reduce(“”,(a,b)->b+” “+a));
Avinash Ujjwal

I have written this code and tried to keep it simple. I think I have considered space and time complexity as well at same time. What’s your input about it, please let me know.

public static String stringReverse(String string) {
if(string == null) {
return “”;
}

StringBuilder stringBuilder = new StringBuilder(“”);

for(int i = string.length() – 1; i >= 0; i–) {
if(Character.isWhitespace(string.charAt(i))) {
stringBuilder.append(string.substring(i + 1, string.length() – stringBuilder.length())).append(” “);;
}
}

stringBuilder.append(string.substring(0, string.length() – stringBuilder.length()));
return stringBuilder.toString();
}
Satyendra Talatam

public void reverseString() {
String original = “Please reverse me hello world”;
int staticPointer = original.length() – 1;
for (int movingPointer = original.length() – 1; movingPointer >= 0; movingPointer–) {
if (original.charAt(movingPointer) == ‘ ‘) {
int movingPointerTemp = movingPointer + 1;
while (movingPointerTemp <= staticPointer) {
System.out.print(original.charAt(movingPointerTemp));
movingPointerTemp = movingPointerTemp + 1;
}
staticPointer = movingPointer – 1;
System.out.print(" ");
}
}

for (int i = 0; i < staticPointer; i++) {
System.out.print(original.charAt(i));
}
}
NightCoder

private static String reverse(String string) {
// TODO Auto-generated method stub

String[] array = string.split(” “);
int start = 0;
int end = array.length – 1;

while(start < end){
String temp = array[start];
array[start] = array[end];
array[end] = temp;
start++;
end–;
}

String str1 = Arrays.toString(array);
return str1.substring(1,str1.length() – 1).replace(",", "");
}
Noe Alejandro Perez Domínguez

Possible one liner in Scala

def reverseWords(s: String): String = s.split(' ').reverse.mkString(" ")
Amit Chaudhary

This works.
omar farag

I have written this code, and I wonder if using the (“out” String) means allocating extra space?

public static String StringRev(String s) { String out=""; if(s==null) return "";
for(int i=s.length()-1;i>0;i--) { if(Character.isWhitespace(s.charAt(i))) { s+=s.substring(i+1)+" "; s= s.substring(0, i); } } out+= s; return out; }
Christian

How about this…

public static String reverseWordsInString(String str){
String[] arr = str.split(” “);
String newArr = “”;
for(int index = arr.length – 1; index >= 0; –index)
newArr += ” ” + arr[index];
return newArr.substring(1);
}

…where I used the ‘substring’ method to produce an output without space. I’m very new to Java and relatively new to programming so I don’t know how Java’s ‘behind-the-scenes’ work, thus I don’t know if my code is more efficient.
Raghu Mahajan

cant use extra space
Raghu Mahajan

we cant use any extra space
Amit Chaudhary

He means the input string is not always “The sky is blue”. It may be longer than 4 words. In that case, the 4 line code solution by stefan fails.
Samir Vasani

public static void main(String[] args) {
// TODO Auto-generated method stub
String s=”my name is ab be fh”,temp;
String[] sDel=s.split(” “);
int j= sDel.length-1;
for(int i=0 ; i<(sDel.length/2) ; i++){
temp =sDel[i] ;
sDel[i] = sDel[j] ;
sDel[j] = temp;
j–;
}

for(int k=0;k<sDel.length;k++){
System.out.println(sDel[k]);
}
}
Samir Vasani

String s=”my name is ab be fh”,temp;
String[] sDel=s.split(” “);
int j= sDel.length-1;
for(int i=0 ; i<(sDel.length/2) ; i++){
temp =sDel[i] ;
sDel[i] = sDel[j] ;
sDel[j] = temp;
j–;
}

for(int k=0;k<sDel.length;k++){
System.out.println(sDel[k]);
}
Akash Chaturvedi

the sky is not always blue 😀
Guest

In place? Without allocating extra space?
stefan

// 4 line of code not 14
public String reverseWordsInString(String str){

String[] arr = str.split(” “); // the sky is blue
String newArr = arr[3] + ” ” + arr[2] + ” ” + arr[0] +” “+ arr[1] ;
return newArr ;
}

LeetCode – Reverse Words in a String II (Java)

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