Question

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

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:
  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

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;
  }
};

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;
}

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
  }
}

/**
 * @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;
};

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;
  }
}

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;
  }
}