README_EN.md

58. Length of Last Word

中文文档

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

JavaScript

/**
 * @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;
};

TypeScript

function lengthOfLastWord(s: string): number {
    s = s.trimEnd();
    const n = s.length;
    const index = s.lastIndexOf(' ');
    if (index !== -1) {
        return n - index - 1;
    }
    return n;
}

Rust

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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