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58. Length of Last Word

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 <= 104
  • s 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 $O(1)$.

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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
}

  • 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;
}


  • /**
 * @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;
};


  • 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;
    }
}

  • 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;
    }
}

  • 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
    }
}

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