# LeetCode – Length of Last Word (Java)

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. If the last word does not exist, return 0.

Java Solution

We can scan the string from the end. When there is a letter, start counting the number of letters. This should handle cases like " a ".

```public int lengthOfLastWord(String s) { if(s==null || s.length() == 0) return 0;   int result = 0; int len = s.length();   boolean flag = false; for(int i=len-1; i>=0; i--){ char c = s.charAt(i); if((c>='a' && c<='z') || (c>='A' && c<='Z')){ flag = true; result++; }else{ if(flag) return result; } }   return result; }```
Category >> Algorithms >> Interview
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• David Moseby

It handles the former but you are correct it does not handle strings ending with an empty space

• ryanlr

The lastIndexOf() method may not be allowed during the interview. It’s a good idea to ask the interviewer first.

Your code does not handle the cases like ” a ” or “a “.

• David Moseby

``` public static int lengthOfLastWord(String s) { if (s == null) return 0;```

``` ```

``` int i = s.lastIndexOf(' ') + 1; System.out.println(s.length() - i); return s.length() - i; } ```

Isn’t this simpler? it also handles Alphanumerics, hyphens and other special characters if needed.

• neha patwardhan

public int lengthOfLastWord(String s) {
int count = 0;
boolean flag = false;
for(int i=s.length()-1; i>=0;i–){
if(s.charAt(i)== ‘ ‘ && !flag){
continue;
}
if(s.charAt(i)==’ ‘ && flag) break;
count++;
flag=true;
}
return count;
}

• CG

The if statement in the for-loop needs to be updated to include a hyphenated last word.

if((c>=’a’ && c=’A’ && c<='Z') || c == '-')

• Eugene Arnatovich

``` public int solution(String s) { if (s == null) { return 0; } s = s.trim(); if (s.length() == 0) { return 0; } String[] words = s.split("\s"); return words[words.length-1].length(); }```

• James

public class Solution {

public int lengthOfLastWord(String s) {

if (s.length() == 0 || s == null)

return 0;

int total_len = s.length();

int lastWordLen = 0;

for (int i = total_len – 1; i >= 0; i–) {

char c = s.charAt(i);

if ((c >= ‘a’ && c = ‘A’ && c 0)

return lastWordLen;

}

}

return lastWordLen;

}

}

• NB****

``` void FunctionRunner::FindLastWordSize_Test() { string str = "My Last word is Harry"; size_t pos = str.find_last_of(" "); if (pos == string::npos) { cout << "Last word size: " << 0 << "n"; } else { cout << "Last word size: " << str.length() - pos - 1 << "n"; } }```

``` ```

• Vimukthi Weerasiri

public int lengthOfLastWord(String s) {
int lastSpace = 0;
s = s.trim();
for (int i = 0; i < s.length() ; i++) {
if (s.charAt(i) == ' ') lastSpace = i + 1;
}
return s.length() – lastSpace;
}

• Ecch IOni

You just doubled the space used :), but honestly in real world we all would do it because it’s faster to type.

• jensiepoo

public int lengthOfLastWord(String s) {
s = s.trim();
String[] arr = s.split(” “);

if(s.length() == 0){
return 0;
}
else{
return arr[arr.length-1].length();
}
}