In the Leetcode Length of Last Word problem solution Given a string s consisting of words and spaces, return the length of the last word in the string.
A word is a maximal substring consisting of non-space characters only.
Example 1:
Input: s = “Hello World”
Output: 5
Explanation: The last word is “World” with length 5.
Example 2:
Input: s = ” fly me to the moon “
Output: 4
Explanation: The last word is “moon” with length 4.
Example 3:
Input: s = “luffy is still joyboy”
Output: 6
Explanation: The last word is “joyboy” with length 6.
Constraints:
- 1 <= s.length <= 104
- s consists of only English letters and spaces ‘ ‘.
- There will be at least one word in s.
Solution in C Programming
#include <stdio.h>
#include <string.h>
int lengthOfLastWord(char* s) {
// trim the trailing spaces
int p = strlen(s) - 1;
while (p >= 0 && s[p] == ' ') {
p--;
}
// compute the length of last word
int length = 0;
while (p >= 0 && s[p] != ' ') {
p--;
length++;
}
return length;
}Solution in C++ Programming
class Solution {
public:
int lengthOfLastWord(string s) {
int len = 0, i;
for(i=s.length()-1;i>=0;i--) {
if(len && s[i] == ' ') return len;
if(s[i] == ' ') continue;
else len++;
}
return len;
}
};Solution in Java Programming
import java.util.*;
class Solution {
public int lengthOfLastWord(String s) {
StringTokenizer st = new StringTokenizer(s);
String sub = "";
while (st.hasMoreElements()) {
sub = st.nextToken();
}
return sub.length();
}
}
Solution in Python Programming
class Solution:
# @param {string} s
# @return {integer}
def lengthOfLastWord(self, s):
n = len(s)
#print n
if n == 1:
if s[0] == ' ':
return 0
else:
return 1
count = 0
if n==0:
return 0
for i in range(n-1,-1,-1):
print 'kk: ',i,s[i]
if s[i] != ' ':
count += 1
elif s[i] == ' ' and count > 0:
return count
return countSolution in C# Programming
public class Solution {
public int LengthOfLastWord(string s) {
string[] words = s.Trim().Split(' ');
return words[words.Length - 1].ToCharArray().Length;
}
}Also read,