Description
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 <= 104s consists of only English letters and spaces ' '.- There will be at least one word in
s.
Solutions
Solution 1: Reverse Traversal + Two Pointers
We start traversing from the end of the string s, find the first character that is not a space, which is the last character of the last word, and mark the index as i. Then continue to traverse forward, find the first character that is a space, which is the character before the first character of the last word, and mark it as j. Then the length of the last word is i - j.
The time complexity is O(n), where n is the length of the string s. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScriptC#PHP
class Solution:
def lengthOfLastWord(self, s: str) -> int:
i = len(s) - 1
while i >= 0 and s[i] == ' ':
i -= 1
j = i
while j >= 0 and s[j] != ' ':
j -= 1
return i - j(code-box)
class Solution {
public int lengthOfLastWord(String s) {
int i = s.length() - 1;
while (i >= 0 && s.charAt(i) == ' ') {
--i;
}
int j = i;
while (j >= 0 && s.charAt(j) != ' ') {
--j;
}
return i - j;
}
}(code-box)
class Solution {
public:
int lengthOfLastWord(string s) {
int i = s.size() - 1;
while (~i && s[i] == ' ') {
--i;
}
int j = i;
while (~j && s[j] != ' ') {
--j;
}
return i - j;
}
};(code-box)
func lengthOfLastWord(s string) int {
i := len(s) - 1
for i >= 0 && s[i] == ' ' {
i--
}
j := i
for j >= 0 && s[j] != ' ' {
j--
}
return i - j
}(code-box)
function lengthOfLastWord(s: string): number {
let i = s.length - 1;
while (i >= 0 && s[i] === ' ') {
--i;
}
let j = i;
while (j >= 0 && s[j] !== ' ') {
--j;
}
return i - j;
}(code-box)
impl Solution {
pub fn length_of_last_word(s: String) -> i32 {
let s = s.trim_end();
let n = s.len();
for (i, c) in s.char_indices().rev() {
if c == ' ' {
return (n - i - 1) as i32;
}
}
n as i32
}
}(code-box)
/**
* @param {string} s
* @return {number}
*/
var lengthOfLastWord = function (s) {
let i = s.length - 1;
while (i >= 0 && s[i] === ' ') {
--i;
}
let j = i;
while (j >= 0 && s[j] !== ' ') {
--j;
}
return i - j;
};(code-box)
public class Solution {
public int LengthOfLastWord(string s) {
int i = s.Length - 1;
while (i >= 0 && s[i] == ' ') {
--i;
}
int j = i;
while (j >= 0 && s[j] != ' ') {
--j;
}
return i - j;
}
}(code-box)
class Solution {
/**
* @param String $s
* @return Integer
*/
function lengthOfLastWord($s) {
$count = 0;
while ($s[strlen($s) - 1] == ' ') {
$s = substr($s, 0, -1);
}
while (strlen($s) != 0 && $s[strlen($s) - 1] != ' ') {
$count++;
$s = substr($s, 0, -1);
}
return $count;
}
}(code-box)
